Answer:
With what?
Step-by-step explanation:
Which of the following groups of animals could be found in temperate grasslands?
Answer asap what’s the correct answer
All groups of animals could be found in temperate grasslands.
Grasslands occupy about a 1/4 of the earth's surface. Grasslands develop in large part as a result of climate. The two main classifications of grasslands are temperate grassland as well as tropical grassland.The American prairies are home to a variety of animals, including prey animals, wild dogs, buffalo, jackrabbits, wolves, and deer. There are many different animals living in the African veldt, such as gazelles, zebra, and rhinos. Rodents, rats, gazelle, beavers, bears, and many other animals are among the fauna of the steppe.Mammals including wolf, prairie animals, reindeer, mice, coyotes, wolves, and jackrabbits are among the inhabitants of this biome. Starlings, sparrows, owls, raptors, and quails are some of the birds found in this biome. The most common insects are grasshoppers and snakes.
All groups of animals could be found in temperate grasslands.
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The following transactions were completed by Winklevoss Inc., whose fiscal year is the calendar year:
Year 1
July 1 Issued $71,900,000 of 20-year, 6% callable bonds dated July 1, Year 1, at a market (effective) rate of 7%, receiving cash of $64,222,669. Interest is payable semiannually on December 31 and June 30.
Oct. 1 Borrowed $420,000 by issuing a six-year, 5% installment note to Nicks Bank. The note requires annual payments of $82,747, with the first payment occurring on September 30, Year 2.
Dec. 31 Accrued $5,250 of interest on the installment note. The interest is payable on the date of the next installment note payment.
31 Paid the semiannual interest on the bonds. The bond discount amortization of $191,933 is combined with the semiannual interest payment.
Year 2
June 30 Paid the semiannual interest on the bonds. The bond discount amortization of $191,933 is combined with the semiannual interest payment.
Sept. 30 Paid the annual payment on the note, which consisted of interest of $21,000 and principal of $61,747.
Dec. 31 Accrued $4,478 of interest on the installment note. The interest is payable on the date of the next installment note payment.
31 Paid the semiannual interest on the bonds. The bond discount amortization of $191,933 is combined with the semiannual interest payment.
Year 3
June 30 Recorded the redemption of the bonds, which were called at 98. The balance in the bond discount account is $6,909,599 after payment of interest and amortization of discount have been recorded. Record the redemption only.
Sept. 30 Paid the second annual payment on the note, which consisted of interest of $17,913 and principal of $64,834.
Required:
1. Journalize the entries to record the foregoing transactions. Round all amounts to the nearest dollar. Refer to the Chart of Accounts for exact wording of account titles.
2. Indicate the amount of the interest expense in (a) Year 1 and (b) Year 2.
3. Determine the carrying amount of the bonds as of December 31, Year 2.
1. The journal entries recording the transactions for Winkelvoss Inc. are as follows:
Journal Entries:Year 1:
July 1: Debit Cash $64,222,669
Debit Bonds Discount $7,677,331
Credit Bonds Payable $71,900,000
Oct. 1: Debit Cash $420,000
Credit Bank Notes Payable $420,000
Dec. 31 Debit Interest Expense $5,250
Credit Installment Note Interest Payable $5,250
Dec. 31 Debit Interest Expense $2,708,433
Credit Cash $2,516,500
Credit Bond Amortization $191,933
Year 2:
June 30 Debit Interest Expense $2,516,500
Credit Cash $2,324,567
Credit Bond Amortization $191,933
Sept. 30 Debit Interest Expense $15,750
Debit Installment Note Interest Payable $5,250
Debit Notes Payable $61,747
Credit Cash $82,747
Dec. 31 Debit Interest Expense $4,478
Credit Installment Note Interest Payable $4,478
Dec. 31 Debit Interest Expense $2,708,433
Credit Cash $2,516,500
Credit Bond Amortization $191,933
Year 3:
June 30 Debit Bonds Payable $64,990,401
Debit Bond Redemption Premium $5,471,599
Credit Cash $70,462,000
Sept. 30 Debit Interest Expense $13,435
Debit Installment Note Interest Payable $4,478
Debit Notes Payable $64,834
Credit Cash $82,747
2. The amounts of the interest expense in Year 1 and Year 2 are as follows:
Year 1 Year 2
Total interest $2,713,683 $2,72,8661
3. The bond's carrying amount as of December 31, Year 2, is $64,798,468.
Transaction Analysis:Year 1:
July 1: Cash $64,222,669 Bonds Discount $7,677,331 Bonds Payable $71,900,000
Oct. 1: Cash $420,000 Bank Notes Payable $420,000
Dec. 31 Interest Expense $5,250 Installment Note Interest Payable $5,250
Dec. 31 Interest Expense $2,708,433 Cash $2,516,500 Bond Amortization $191,933
Year 2:
June 30 Interest Expense $2,516,500 Cash $2,324,567 Bond Amortization $191,933
Sept. 30 Interest Expense $15,750 Installment Note Interest Payable $5,250 Notes Payable $61,747 Cash $82,747
Dec. 31 Interest Expense $4,478 Installment Note Interest Payable $4,478
Dec. 31 Interest Expense $2,708,433 Cash $2,516,500 Bond Amortization $191,933
Year 3:
June 30: Bonds Payable $64,990,401 Bond Redemption Premium $5,471,599 Cash $70,462,000
Sept. 30 Interest Expense $13,435 Installment Note Interest Payable $4,478 Notes Payable $64,834 Cash $82,747
Interest Expenses:Year 1 Year 2
Bonds Payable $2,708,433 $2,708,433
Notes Payable $5,250 $15,750
Notes Payable $4,478
Total interest $2,713,683 $2,72,8661
Carrying amount of the bonds December 31, Year 2:July 1: $64,222,669
Dec. 31 Bond Amortization $191,933
June 30 Bond Amortization $191,933
Dec. 31 Bond Amortization $191,933
Dec. 31 Year 2 Carrying Amont = $64,798,468
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Look at the following problem, find the mistake.Solve for x. 6x − 3x + 14 = 176x − 3x + 14 = 17 9x + 14 = 17 - 14 -14 9x = 3 /9 /9 x = 1/3Solve for x. 6x − 3x + 14 = 17
2. Daniel is baking pie for his each of his class periods. His equation is P-1c. P
represents the number of Pies and c represents the number of classes.
What is this equation saying?
Stary
What are we comparing?
Write equation as ratio.
Here, we are comparing Daniels's class periods and pie. They equal each other.
The required ratio is 1:1.
What are we comparing here?
Given,
P-1c
Solution:
P-1c, where c is class periods and p is pies.
P-1c = 0
1p = 1c
Which means,
He bakes 1 pie per class period.
His total number of pies equals his total number of class periods.
The equation can be written in ratios
1p = 1c i.e., they are in 1:1 ratio.
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The line y − 14 = 6(x − 2.5) represents Barry’s profit, y, from selling x painting, aftser buying some canvas. What was the cost of the canvas?
The cost of the canvas will be 6x.
What is Equation of line?
The equation of line in point-slope form passing through the points
(x₁ , y₁) and (x₂, y₂) with slope m is defined as;
⇒ y - y₁ = m (x - x₁)
Where, m = (y₂ - y₁) / (x₂ - x₁)
Given that;
The equation of line is,
⇒ y - 14 = 6 (x - 2.5)
Now,
The line represents Barry’s profit, y, from selling x painting, after buying some canvas.
Here, The equation of line is,
⇒ y - 14 = 6 (x - 2.5)
Simplify the equation as;
⇒ y - 14 = 6 (x - 2.5)
⇒ y - 14 = 6x - 15
⇒ y = 6x - 15 + 14
⇒ y = 6x - 1
Since, Barry’s profit, y, from selling x painting, after buying some canvas.
Thus, The cost of the canvas 6x.
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4 (-2,6)
B (4,6)
C(-2,3)
D(4,3)
Answer:
answer is c-(2,3)
Step-by-step explanation:
......................m
Can you help me with this
Given:
[tex]\begin{gathered} y=x+4 \\ y=x+2 \\ x=1 \\ x=4 \end{gathered}[/tex]Required:
To find the volume of the solid by using washer method.
Explanation:
Volume formula of Washer method is,
[tex]V=\int_a^b\pi[(f(x)^2-g(x)^2]dx[/tex]Therefore,
[tex]\begin{gathered} V=\int_1^4\pi[(x+4)^2-(x+2)^2]dx \\ \\ =\int_1^4\pi[x^2+16+8x-x^2-4-4x]dx \\ \\ =\int_1^4\pi[4x+12]dx \\ \\ =\pi[\frac{4x^2}{2}+12x]_1^4 \\ \\ =\pi[\frac{64}{2}+48-\frac{4}{2}-12] \\ \\ =\pi[32+48-2-12] \\ \\ =66\pi \end{gathered}[/tex]Final Answer:
Volume is
[tex]66\pi[/tex]A man has 32 coins in his pocket, all of which are dimes and quarters. If the total value of his change is 545 cents, how many dimes does he have?
A dime is worth 10 cents and a quarter is worth 25 cents.
Let D be the number of dimes and Q be the number of quarters.
Since the total amount of coins in the pocket is 32, then:
[tex]D+Q=32[/tex]On the other hand, the total value of D dimes is 10D, while the total value of Q dimes is 25Q. Then, the total value of D dimes and Q cuarters is 10D+25Q, which must be equal to 545. Then:
[tex]10D+25Q=545[/tex]Notice that we have found a 2x2 system of equations:
[tex]\begin{gathered} D+Q=32 \\ 10D+25Q=545 \end{gathered}[/tex]Solve the system using the substitution method. To do so, isolate D from the first equation and replace the expression for D into the second equation to obtain a single equation in terms of Q:
[tex]\begin{gathered} D+Q=32 \\ \Rightarrow D=32-Q \\ \\ 10D+25Q=545 \\ \Rightarrow10(32-Q)+25Q=545 \\ \Rightarrow320-10Q+25Q=545 \\ \Rightarrow25Q-10Q=545-320 \\ \Rightarrow15Q=225 \\ \Rightarrow Q=\frac{225}{15} \\ \\ \therefore Q=15 \end{gathered}[/tex]Replace back Q=15 into the expression for D to find the amount of dimes:
[tex]\begin{gathered} D=32-Q \\ =32-15 \\ =17 \end{gathered}[/tex]Therefore, the amount of dimes that the man has is 17.
The population of the world in 1987 was 5 billion and the annual growth rate was estimated at 2 percent per year. Assuming that the world population follows an exponential growth model, find the projected world population in 1992.
Answer:
5.52 billion
Step-by-step explanation:
2% s .02 in decimal from '87 to '92 is 5 years ( the exponent in the following:
5 (1.02)^5 = 5.52 billion
Answer:
5.52 billion
Step-by-step explanation:
i got this my doing math
q + ad = m solve for a
Answer:
a = [tex]\frac{m-q}{d}[/tex]
Step-by-step explanation:
q + ad = m ( subtract q from both sides )
ad = m - q ( isolate a by dividing both sides by d )
a = [tex]\frac{m-q}{d}[/tex]
Answer:
a = (m - q)/d
Step-by-step explanation:
Given that,
→ q + ad = m
Now solving for the value of a,
→ q + ad = m
→ ad = m - q
→ a = (m - q)/d
Hence, the value is (m - q)/d.
The width of a rectangle is 2 its length. The perimeter of the rectangle is 540 ft. What is the length, in feet, of the rectangle?
To solve this problem we can write one equation for each condition so:
the width of a rectangle is 2 its length so:
[tex]W=2L[/tex]and one for the perimeter so:
[tex]2W+2L=540[/tex]Now we replace the first equation into the second one so:
[tex]\begin{gathered} 2(2L)+2L=540 \\ 4L+2L=540 \\ 6L=540 \\ L=\frac{540}{6} \\ L=90 \end{gathered}[/tex]So the length of the rectangle is 90 ft
Help needed! please help math
PLEASEEE!!! Math 8th grade
Total gallons of water used in the park after the ride is installed
= 9.51 * 10^5 GALLONS.
What is exponent addition?Exponent addition is the process of multiplying a number by its exponents or powers, whether or not the base is the same. Exponents, which show how many times a number can be multiplied by itself, are also known as the power of numbers. As an illustration, 3^2 = 3*3, where 3 is the base and 2 the exponent.
How to add exponents?When the base and exponents are the same, exponents can be added. Sometimes the base and exponent will differ, but adding can still be done for certain formulas. Let's examine the procedures for adding exponents.
Step 1: Verify that the expression's terms all have the same base and exponents. 22 + 22, as an illustration. As can be seen, the exponent and base are both 2.
Step 2: Calculate the expression using individual terms if the base and exponent are different. for instance, 53 plus 42. Exponents and bases are different.
Step 3: Add the results together.
As per the question:water used in the park initially = 8.6*10^5 gallons
water used by the additional ride added = 9.1 *10^4 gallons
=0.91*10^5 gallons
Total gallons of water used in the park after the new ride is installed
= (8.6*10^5 + 0.91*10^5) gallons
= (8.6+0.91) *10^5 gallons
= 9.51* 10^5 gallons
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How do you write – 7.83 as a fraction into simples form?
7.83 can be written as a fraction in simplified form as [tex]7 \frac{80}{100}[/tex].
A fraction, which is expressed in the form p/q, where p and q are both integers, denotes a portion of a whole. Here, we'll explain the steps involved in converting 7.83 decimal numbers to fraction form and mixed numbers. In order to convert 7.83 to a fraction, do the following:
Write the decimal number split by one first as follows:
7.83/1
Since the numerator has two digits after the decimal point, we must multiply the numerator and denominator by 102 = 100, removing the decimal point.
7.83 × 100/1 × 100 = 783/100
We can also express 7.83 as a mixed number because the improper fraction is caused by the numerator being bigger than the denominator; as a result, 783/100 is equal to:
[tex]7 \frac{80}{100}[/tex]
Therefore, 7.83 is 783/100 as a fraction, and its mixed number form is [tex]7 \frac{80}{100}[/tex].
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Which equation or inequality represents the following description? & Multiply the quantity 2 more than a number by 8. The result is at least 12 times the number subtracted from 76.A. 8(2n) < 76 - 12B. 8n + 2 = 12n - 76C. 8(n + 2) > 76 - 12nD. 8(2n) = 76 – 12n
We can translate the wording part as follows:
1. Multiply the quantity 2 more than a number by 8:
8 * (2 + n)
2. The result is at least: it means that the value is the value in question or values greater than this number, therefore, we have an inequality here ( >=).
3. The result is at least 12 times the number subtracted from 76:
76 - 12n
Therefore, the inequality that represents the description is:
[tex]8\cdot(2+n)\ge76-12n[/tex]Then, the correct option is C.
Giving a test to a group of students, the grades and gender are summarized belowIf one student is chosen at random,Find the probability that the student was male OR got an "B".
ANSWER:
0.6875
STEP-BY-STEP EXPLANATION:
The first thing is to calculate the probability that the gender is male and also the probability that the grade is B, separately, like this:
[tex]\begin{gathered} P(\text{male})=\frac{42}{80} \\ P(B)=\frac{19}{80} \end{gathered}[/tex]Therefore, since it is the probability that it is male OR got and "B", it is the union of both events, and it would look like this:
[tex]P(\text{male}\cup B)=P(male)+P(B)-P(male\cap B)[/tex]Now, the intersection of both events would be the probability that he is a man and gets a B, it would look like this:
[tex]P(male\cap B)=\frac{6}{80}[/tex]We replace to be able to calculate the union, like this:
[tex]\begin{gathered} P(\text{male}\cup B)=P(male)+P(B)-P(male\cap B) \\ P(\text{male}\cup B)=\frac{42}{80}+\frac{19}{80}-\frac{6}{80} \\ P(\text{male}\cup B)=\frac{55}{80}=\frac{11}{16}=0.6875 \end{gathered}[/tex]The probability that the student was male OR got an "B" is 0.6875
Find the real solutions, if any, of the following equation. Use the quadratic formula.
EXPLANATION
Given the equation 6x^2 = 5x
We can apply the following procedure :
Subtracting -5x to both sides:
[tex]6x^2-5x=0[/tex]Apply exponent rule:
[tex]a^{(b+c)}=a^ba^c[/tex][tex]x^2=\times[/tex][tex]=6\times-5x[/tex]Factor out common term x:
[tex]x\mleft(6x-5\mright)=0[/tex][tex]\mathrm{Using\: the\: Zero\: Factor\: Principle\colon\quad \: If}\: ab=0\: \mathrm{then}\: a=0\: \mathrm{or}\: b=0[/tex][tex]x=0\quad \mathrm{or}\quad \: 6x-5=0[/tex][tex]\mathrm{Add\: }5\mathrm{\: to\: both\: sides}\text{ from 6x -5=0}[/tex][tex]6x-5+5=0+5[/tex][tex]Simplify\colon[/tex][tex]6x=5[/tex][tex]\mathrm{Divide\: both\: sides\: by\: }6[/tex][tex]\frac{6x}{6}=\frac{5}{6}[/tex][tex]Simplify\colon[/tex][tex]x=\frac{5}{6}[/tex][tex]\mathrm{The\: solutions\: to\: the\: quadratic\: equation\: are\colon}[/tex][tex]x=0,\: x=\frac{5}{6}[/tex]Hence, the solution set is as follows:
{0 , 5/6}
What principal would you need to invest at a rate of 4% to earn $500 in 6 months? Round your answer to the nearest cent.
From the information available, we have the following;
Rate = 4% (0.04)
Time = 0.5 (half a year/6 months)
Interest = 500
Principal = ???
Therefore, we would substitute these into the simple interest formula as shown below;
[tex]\begin{gathered} I=P\times R\times T \\ \text{Make P the ubject of the equation;} \\ \text{Divide both sides by R x T and you'll have;} \\ \frac{I}{R\times T}=\frac{P\times R\times T}{R\times T} \\ \frac{I}{R\times T}=P \\ \frac{500}{0.04\times0.5}=P \\ \frac{500}{0.02}=P \\ P=25000 \end{gathered}[/tex]ANSWER:
The principal to be invested therefore is $25,000
Can you please help me out with a question
There are 360 degrees in a circle.
We can write:
[tex]Arc\text{LAM}+Arc\text{MBL}=360\degree[/tex]Given, Arc LAM = 256°, we can find Arc MBL:
[tex]\begin{gathered} Arc\text{LAM}+Arc\text{MBL}=360\degree \\ 256+\text{ArcMBL}=360 \\ \text{ArcMBL}=360-256 \\ \text{ArcMBL}=104 \end{gathered}[/tex]The central angle that subtends Arc MBL also measures 104 degrees.
[tex]\angle\text{MPL}=104\degree[/tex]We also know,
[tex]\angle\text{MPL}=\angle\text{MPB}+\angle\text{BPL}[/tex]Angle MPB and Angle BPL are equal, so we have:
[tex]\begin{gathered} \angle\text{MPL}=\angle\text{MPB}+\angle\text{BPL} \\ 104=2\angle\text{BPL} \\ \angle\text{BPL}=\frac{104}{2} \\ \therefore\angle\text{BPL}=52\degree \end{gathered}[/tex]Now,
Arc LB subtends the central angle BPL, so they are same in measure.
Thus,
[tex]\text{ArcLB}=52\degree[/tex]Please help! Functions and Relations. The function h(x) is a transformation of the square root function, f(x)= square root of x. What function is h(x)? Thanks!
In general, given a function g(x), a horizontal shift is given by the transformation below
[tex]\begin{gathered} g(x)\rightarrow g(x-a) \\ a>0\rightarrow\text{ a units to the right} \\ a<0\rightarrow\text{ a units to the left} \end{gathered}[/tex]Thus, in our case, notice that the graph of h(x) is that of f(x) shifted 1 unit to the left; then,
[tex]h(x)=\sqrt{x+1}[/tex]The answer is option C.What is the sum of the two odd numerals following 26 in the following sequence?
2,3,5,6,7,9,10,11,13…
The two odd numbers are 11 and 15. The sum of the odd numbers in the sequence will be 26.
What is a sequence?It is defined as the systematic way of representing the data that follows a certain rule of arithmetic.
Divergent sequences are those in which the terms never stabilize; instead, they constantly increase or decrease as n approaches infinity, approaching either infinity or -infinity.
It is given that,
The sequence is,
2,3,5,6,7,9,10,11,13…
As we know the given series is an arithmetic progression with a common difference is 2.
The sequence is further obtained as,
2,3,5,6,7,9,10,11,13, 15
From the given condition the sum of the numerals has to be 26.After applying a lot of addition to the sequence we get only two odd numbers 11 and 15 which will give the sum of 26 as,
= 11 + 15
=26
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Andy Lee is the punter for the San Francisco 49ers. He had a stellar 2011 season with an average punt length of 50.9 yards with a standard deviation of 3.5. His punt distance follows a normal distribution. Andy Lee's first punt of the season was 66 yards. what is the z-score for this punt?A. -2.4B. 2.4C. 4.3D.5.5
The z-score of a normal distribution is defined by
[tex]Z=\frac{x-\mu}{\sigma}[/tex]where
[tex]\begin{gathered} x\text{ is the score we want to find } \\ \mu\text{ is the mean value of the distribution} \\ \sigma\text{ is the standard deviation of the distribution} \end{gathered}[/tex]Then, using the formula for the z-score in our problem we have
[tex]Z=\frac{66-50.9}{3.5}=\frac{15.1}{3.5}=4.3[/tex]Then the z score for that punt is 4.3.
(C) 0.845 ÷ 5 I need help explaining the answer
The given expression is
[tex]\frac{0.845}{5}[/tex]As long division, we solve it as follows
So, the answer is 0.169.Observe that we have to add a zero first in the quotient in order to transform 0.845 into 845, then we divide it by 5 as a normal long division.
What is the approximate circumference of a circle with a radius of 13 feet?
ANSWER
81.68 ft
EXPLANATION
The circumference of a circle = 2 * pi * r
[tex]\text{circumference = 2 }\Pi\text{ r = 2}\times\frac{22}{7}\times13\text{ = 81.68 ft}[/tex]The result of RREF produced the following matrix. What is the solution of the system? Choose the best answer.
Using the augmented matrix, we find that:
The first system has no solutions.
The second system has infinite solutions, given by: (-3t, 0, t).
Augmented Matrix:
An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation and each column represents all the coefficients for a single variable.
Given:
The matrix is,
[tex]\begin{bmatrix}1 & 0 & 3&| & 0 \\ 0& 1 & 1 & | & 5 \\0 & 0 & 0 & | & 1\end{bmatrix}[/tex]
Here we need to find the solution of the system.
While we looking at the first system, at the third row of the matrix, the relation is written as,
0t=1
This one is not possible, therefore, there are no solutions for the first system.
Now when we looking at the second system, the third line has the value of,
0t=1
This refers that there are infinite solutions, therefore here t is the free parameter.
From the second line, y = 5.
From the first,
=> x + 3t
=> x = -3t
The solutions are: (-3t, 0, t).
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HELP ME OUT PLEASE!!!!!!!!
The distance from Fort Worth to Austin is about 190 miles. Which of the following equations describes the remaining distance to Austin for a car that travels from Fort Worth at a rate of 65 miles per hour?
The linear function that gives the remaining distance to Austin for a car that travels from Fort Worth at a rate of 65 miles per hour is:
y = 190 - 65x.
What is a linear function?The general format of a linear function is defined as follows:
y = mx + b.
In which:
m is the slope, representing the rate of change of y relative to x.b is the intercept, which is the initial value of the function.The parameters in the context of this problem are as follows:
b = 190, which is the initial distance between the two cities.m = -65, as due to the velocity, the distance decays by 65 each hour.Hence the equation is:
y = 190 - 65x.
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Can’t find the correct answer pls help
a)What is the value of y? y= 6
b)The value of RS is 57 and ST is 16.
a) How is the value of y calculated?
From the number line,
RS = 9y +3
ST= 2y + 4
RT = 73
Rewriting as equation,
RT = RS + ST
73 = 9y +3 + 2y + 4
73 = 11y + 7
11 y = 73 - 7
11y = 66
y = 66/11
y = 6
b) What is the value of RS and ST ?
Considering the equation,
RS = 9y + 3
=9(6) +3
=54 +3
RS = 57
Considering the equation,
ST = 2y + 4
=2(6) +4
= 12 +4
ST = 16
What is an equation?
An equation is a mathematical statement that demonstrates the equality of two mathematical expressions. More than one variable may be present in a linear equation. An equation is said to be linear if the maximum power of the variable is consistently 1. Another name for linear equation is a one-degree equation.To learn more about equations , refer:
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DG and EG are tangent to circle C and circle F. The points of tangency are A, B, D, and E. if M
From the question and the given diagram, we were told that:
DG and EG are tangent to circle C and circle F.
The point of tangency are A, B, D, and E.
If M
We are to find m
In solving this, we will have to need or consider the similarity theorem.
Its says that if corresponding angles are congruent, then their angles are similar.
It in essence states that, C
The product of -1/a≠ 0, and it's reciprocal is?
A. a/2
B. 2/a
C. -2/a
D. None are correct
The product of -1/a and its reciprocal will be equal to unity or 1.
What is the meaning of reciprocal of a number?For any number [a], the reciprocal will be given by [1/a] such that -
a x 1/a = 1
Given is the following expression -
A = -1 / a
From the definition of the reciprocal of a number, the reciprocal of the number -
A = -1 / a
will be
1/A = 1/(-1/a) = -a
Now, the product of the number and its reciprocal will be -
A x 1/A = -1/a x -a = 1
A x 1/A = 1
Therefore, the product of -1/a and its reciprocal will be equal to unity or 1.
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Convert to an algebraic expression and simplify, if y stands for the unknown number: Sixteen more than five times a number.
Here, we are told y stands for the unknown number: Sixteen more than five times a number.
Converting to an algebraic expression, we have:
16 + 5y
Here, y is the unknown number.
16 + 5y
Which equation represents a line that is perpendicular to the graph of y = 2/5x − 1?
The equation of the line y=-5/2x-4 is perpendicular to the given equation of the line.
Given that,
The given equation of the line is y=2/5x-1.
We have to find the equation of the line that is perpendicular to the given equation of the line.
We know Equation of the line formula,
y=mx+c
Here,
m denotes the line's slope, and c the y-intercept.
Take the equation of the line y=2/5x-1
m is 2/5
The perpendicular line is m is -5/2
We get the perpendicular line is y=-5/2x-4.
Therefore, The equation of the line y=-5/2x-4 is perpendicular to the given equation of the line.
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