1) Let's begin by writing down the equations we need.
Freshmen
y=264+29x Note that $29 depends on the number of rolls
Sophomore
y= 241+30x
2) Now that we know the equations that describe those expenses, let's apply the Substitution Method
Let's plug into the 1st equation the value of y.
[tex]undefined[/tex]A football that has a regular price of $18.85 is on sale for 15% off. What is the sale price?
Regular price = $18.85
discount = 15%
18.85 -------------------- 100%
x --------------------- 15%
x = (15 x 18.85) / 100
x = 282.75/100
x = 2.8275
Sale price = 18.85 - 2.8275
Sale price = $16.02
17.graph the function. state the domain and range f(x) =[x-2]19. determine wether each equation is a linear equation. circle yes or no. If yes write the equation in standard form
Solution
17)
Write the function
[tex][/tex]The length of a rectangle is three inches more than the width. the area of the rectangle is 180 inches. find the width of the rectangle
We have a rectangle with length L that is 3 inches more than the width W. Then we can write this as:
[tex]L=W+3[/tex]The area of the rectangle is 180 square inches.
We have to find the width W.
As the area is equal to the product of the length and the width, we can write this equation and solve for W as:
[tex]\begin{gathered} A=180 \\ L\cdot W=180 \\ (W+3)\cdot W=180 \\ W^2+3W=180 \\ W^2+3W-180=0 \end{gathered}[/tex]We have a quadratic equation. The roots of this equation will be the mathematical solutions.
We can find the roots using the quadratic formula:
[tex]\begin{gathered} W=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ W=\frac{-3\pm\sqrt[]{3^2-4\cdot1\cdot(-180)}}{2\cdot1} \\ W=\frac{-3\pm\sqrt[]{9+720}}{2} \\ W=\frac{-3\pm\sqrt[]{729}}{2} \\ W=\frac{-3\pm27}{2} \\ W_1=\frac{-3-27}{2}=-\frac{30}{2}=-15 \\ W_2=\frac{-3+27}{2}=\frac{24}{2}=12 \end{gathered}[/tex]The solutions are W = -15 and W = 12.
The first one is not valid, as W has to be greater than 0.
Then, the solution to our problem is W = 12 in.
Answer: the width is W = 12 inches.
Use the tangent to find the length of side A C. Express your answer to the nearest tenth. 27° 7.0 The length of side AC is approximately units.
From the basic knowledge of trignometry,
tan 27 = x /7.0
cross-multiply
x = 7.0 x tan 27
x = 7.0 x 0.5095
x = 3.5665 = 3.6 units ( nearest tenth)
Translate the following phrase into a math expression whichever is appropriate use x to represent the unknown number The difference between eight and a number
We have the next phrase
"The difference between eight and a number "
Difference means subtraction therefore the expression will be
[tex]undefined[/tex]On a particular day, the amount of untreated water coming into the plant can be modeled by f(t) = 100 + 30cos(t/6) where t is in hours since midnight and f(t) represents thousands of gallons of water. The amount of treated water at any given time, t, can be modeled by g(t) = 30e^cos(/2)a) Define a new function, a(t), that would represent the amount of untreated water inside the plant, at any given time, t.b) Find a′ (t).c) Determine the critical values of this function over the interval [0, 24).d) Determine whether the critical values represent local maximums or minimums.e) Determine the maximum and minimum amount of untreated water in the plant for the day.
The amount of untreated water coming into a plant can be modeled by the following function:
[tex]f(t)=100+30\cos\frac{t}{6}[/tex]The amount of water treated by that same plant is given by:
[tex]g(t)=30e^{\cos\frac{t}{2}}[/tex]In part (a) we must define the function a(t) that represents the amount of untreated water inside the plant, at any given time, t. This quantity is equal to the amount of untreated water entering the plant less the amount of water treated by it at the same time. In summary we have:
[tex]a(t)=f(t)-g(t)=100+30\cos\frac{t}{6}-30e^{\cos\frac{t}{2}}[/tex]In part (b) we have to find its derivative. In order to do that let's remember how to derivate a cosine and an exponential expression:
[tex]\begin{gathered} h(t)=\cos(k(t))\Rightarrow h^{\prime}(t)=-\sin(k(t))\cdot k^{\prime}(t) \\ b(t)=e^{c(t)}\Rightarrow b^{\prime}(t)=e^{c(t)}\cdot c^{\prime}(t) \end{gathered}[/tex]Using these properties we can find a'(t):
[tex]\begin{gathered} a^{\prime}(t)=f^{\prime}(t)-g^{\prime}(t)=(100+30\cos\frac{t}{6})^{\prime}-(30e^{\cos\frac{t}{2}})^{\prime} \\ a^{\prime}(t)=-\frac{30}{6}\sin\frac{t}{6}-30e^{\cos\frac{t}{2}}\cdot(-\frac{1}{2}\sin\frac{t}{2}) \\ a^{\prime}(t)=-5\sin\frac{t}{6}+15e^{\cos\frac{t}{2}}\sin\frac{t}{2} \end{gathered}[/tex]In part (c) we must find the critical points of the function over [0,24). The critical values are t-values where the derivative of the function is not defined or where it is 0. Then we must solve a'(t)=0:
[tex]a^{\prime}(t)=0=-5\sin(\frac{t}{6})+15e^{\cos(t\/2)}\sin(\frac{t}{2})[/tex]Write the algebraic rule that maps the pre-image (solid) onto the image (dashed).
We need to identify the points of the pre-image and image and compare them.
So, points of the pre-image are:
(-6, 8), (-3, 6), (-9,4), (-6,4), (-2,3) and (-3,0)
At the same way, point of the image are:
(-8, -6), (-6, -3), (-4, -9), (-4, -6), (-3, -2) and (0,-3)
Now we can see that if the pre-image point is (x,y) the image point is (-y, x)
Answer: the algebraic rule is that every point (x,y) is moved to point (-y, x)
For example, (-6,8) is moved to point (-8, -6).
13. Consider the function g graphed below. For whatvalues of Xo does lim g(x) exist?
We will say that the limit
[tex]\lim _{x\to xo}f(x)=a[/tex]exists if we reach the same value (a) regardless of the direction we approach x_0
So, in the case of the graph, the function is evidently continuous everywhere except for two points in which we need to determine whether the function is continuous there or not.
Notice that x->-4 is one of those particular points.
We have that:
[tex]\begin{gathered} \lim _{x-\to-4}f(x)=6 \\ \text{and} \\ \lim _{x+\to-4}f(x)=4 \end{gathered}[/tex]Then, we obtain the value 6 if we approach x= -4 from the left and we get 4 if we move towards x= -4 from the right. The two values are not the same, therefore the limit does not exist there.
As for x-> 2
Notice that
[tex]\lim _{x-\to2}f(x)=2=\lim _{x+\to2}f(x)[/tex]Therefore, the limit exists and it is equal to 2 even though the function has a different value there.
Therefore, the limit exists for any point in the set:
[tex]x_o\in\mathfrak{\Re }-\mleft\lbrace-4\mright\rbrace[/tex]I like to know the steps to this type of question please and thank you
In this problem they are telling you that the mean is 64 ounces and that the standar derivation is 2.4 onces so:
To find the maximun value that can have Mu we have tu sum the standart derivation to the mean so:
[tex]\text{maximun}=64+2.4=66.4[/tex]and to calculate the minumun value that can have mu we rest the standar derivation from the mean so:
[tex]\text{minumun}=64-2.4=61.6[/tex]so the solution is:
[tex]61.6<\mu<66.4[/tex]If sin theta=3 divided by square root 15 and angle theta is in Quadrant I, what is the exact value of tan 2 theta insimplest radical form?
The given information is:
[tex]\sin \theta=\frac{3}{\sqrt[]{15}}[/tex]And angle theta is in quadrant I, then know that:
[tex]\sin (x)=\frac{opposite}{\text{hypotenuse}}=\frac{y}{r}[/tex]Then y=3 and r=square root (15)
By the Pythagorean theorem we can find x:
[tex]\begin{gathered} r^2=x^2+y^2 \\ x^2=r^2-y^2 \\ x=\sqrt[]{r^2-y^2} \\ x=\sqrt[]{(\sqrt[]{15})^2-3^2} \\ x=\sqrt[]{15-9} \\ x=\sqrt[]{6} \end{gathered}[/tex]And the tangent is:
[tex]\begin{gathered} \tan (x)=\frac{y}{x} \\ \tan \theta=\frac{3}{\sqrt[]{6}} \end{gathered}[/tex]Thus, tan 2theta:
[tex]\begin{gathered} \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta} \\ \tan 2\theta=\frac{2\frac{3}{\sqrt[]{6}}}{1-(\frac{3}{\sqrt[]{6}})^2} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{1-\frac{9}{6}} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{-\frac{3}{6}} \\ \tan 2\theta=\frac{\frac{6}{\sqrt[]{6}}}{-\frac{1}{2}} \\ \tan 2\theta=\frac{6\cdot2}{-1\cdot\sqrt[]{6}} \\ \tan 2\theta=\frac{12}{-\sqrt[]{6}} \\ \tan 2\theta=-\frac{12}{\sqrt[]{6}} \end{gathered}[/tex]Then the exact value of tan 2theta in simplest radical form is -12/square root(6)
Separate the trapezoid into figures whose areas you can find. Label the dimensions. 6 What is the area of the trapezoid? Show your work.
Given
Graph
Procedure
5) The figure can be separated into a triangle and a rectangle.
The trapezoid can be divided into two figures. A triangle and a rectangle.
6) Are of the trapezoid
[tex]\begin{gathered} A_T=\frac{1}{2}bh \\ A_T=\frac{1}{2}\cdot2\cdot1.5 \\ A_T=1.5\text{ m}^2 \end{gathered}[/tex][tex]\begin{gathered} A_b=w\cdot l \\ A_b=3\cdot1.5 \\ A_b=4.5m^2 \end{gathered}[/tex]Total area
[tex]\begin{gathered} A=A_T+A_b \\ A=1.5+4.5 \\ A=6m^2 \end{gathered}[/tex]It is estimated that about 7,217 African cheetahs are living in the wild today and thepopulation is expected to decline at a rate of 8.4% per year. Predict the number ofAfrican cheetahs living in the wild in 23 years. Round answer to a whole number.
Answer: 959
Explanation:
Step 1. The information that we have is:
The initial or actual amount of African cheetahs is 7,217, we will call this 'a':
[tex]a=7,217[/tex]The rate of decline for the population is 8.4% per year, this will be 'r':
[tex]r=8.4[/tex]Converting this amount from percentage to decimal number:
[tex]\begin{gathered} r=8.4/100 \\ r=0.084 \end{gathered}[/tex]And we require to find the number of cheetahs in 23 years, this will be the time 't':
[tex]t=23[/tex]Step 2. This problem is an example of exponential decay, and the formula to use is:
[tex]y=a(1-r)^t[/tex]Where 'y' represents the final amount after time t.
Step 3. Substituting the known values:
[tex]y=7,217(1-0.084)^{23}[/tex]Step 4. Solving the operations:
[tex]y=7,217(0.916)^{23}[/tex][tex]y=7,217(0.132921702)[/tex]The result is:
[tex]y=959.296[/tex]Rounding this result to the nearest whole number:
[tex]y=959[/tex]The prediction is 959 African cheetahs.
Answer: 959
There are twelve doughnuts in a dozen. How many doughnuts are there in ten dozen?On the double number line below, fill in the given values, then use multiplication ordivision to find the missing value:dozensdoughnutsThere aredoughnuts in 10 dozen.
Given:
12 doughnuts in a dozen
Find how many doughnuts are there in 10 dozens.
To solve for this, let us first take a look at the number line provided. From the problem, it is stated that there are 12 doughnuts in 1 dozen. Therefore, 12 doughnuts = 1 dozen
Next, to find how many doughnuts are there in 10 dozens, we will just multiply 12 by 10.
[tex]10dozen\times\frac{12doughnuts}{1dozen}[/tex][tex]=120doughnuts[/tex]Therefore, we can conclude that there are 120 doughnuts in 10 dozen.
If 39% of the people in a community use the emergency room at a hospital in oneyear, find these probabilities for a sample of 10 people. a) At most 4 used the emergency room.b) At least 5 used the emergency room.
The probability of success p is given by:
[tex]p=0.39[/tex]Recall the binomial probability distribution formula:
[tex]Pr(x)=\begin{pmatrix}n \\ x\end{pmatrix}p^x(1-p)^{n-x}[/tex]The probability of at most 4 is given by:
[tex]Pr(0\leq x\leq4)=P(0)+P(1)+P(2)+P(3)+P(4)[/tex]The probability Pr(0) is given by:
[tex]Pr(0)=\begin{pmatrix}10 \\ 0\end{pmatrix}0.39^0(1-0.39)^{10-0}=0.0071334291166288[/tex]Therefore,
[tex]Pr(0\leq x\leq4)=0.65796153140997\approx0.6580[/tex]Therefore the probability of at most 4 is approximately 0.6580
Using a similar procedure, it follows that the probability of at least 5 is approximately 0.3420
What is 0.42857142857 as a fraction? Express the result in simplest form.
To convert a decimal to a fraction, place the decimal number over its place value.
For instance, in 0.6, the six is in the tenths place, so we place 6 over 10 to create the equivalent fraction,
[tex]\frac{6}{10}[/tex]Similarly, in 0.42857142857 we have 11 places next to the decimal point. Then, we have
[tex]0.42857142857=\frac{42857142857}{100,000,000,000}[/tex]A fraction is in simplest form when the top and bottom cannot be any smaller, while still being whole numbers. In our case, if we divide by 5 both the numerator and denominador, we obtain
[tex]\frac{42857142857/5}{100,000,000,000/5}=\frac{857142857}{20000000000}[/tex]since 5 is the greatest commun factor, this is the simplified answer.
Screen-printing a batch of shirts requires 1 minute per shirt in addition to 5 minutes of initial set-up time. If it takes 44 minutes to screen-print a batch of shirts, how many shirts are in the batch?
The problem is a linear equation
of the form y= ax + b
because every shirt requires 1 minute, that means a=1
and because there are 5 minutes more,that means b=5
so the equation is y = x + 5
which value is y? The problem says its 44 minutes, all the amount of time
so we have 44 = x + 5
List the side of triangle DEF in order from shortest to longest
Apply the Angles Sum Property to the triangle DEF,
[tex]\begin{gathered} \angle D+\angle E+\angle F=180^{\circ} \\ 31^{\circ}+112^{\circ}+\angle F=180^{\circ} \\ 143^{\circ}+\angle F=180^{\circ} \\ \angle F=180^{\circ}-143^{\circ} \\ \angle F=37^{\circ} \end{gathered}[/tex]According to the Sine Rule, the length of side is proportional to the angles measure opposite to it. So the shortest side will be opposite to shortest angle, and the largest side will be opposite to the largest angle.
It is observed that,
[tex]DThis follows that,[tex]EFThus, the order of sides from shortest to longest is EF, DE, DF.O POLYNOMIAL AND RATIONAL FUNCTIONSFinding horizontal and vertical asymptotes of a rational function.
Explanation
Giving the function
[tex]f(x)=\frac{x^2+2x-3}{-x^2-4}[/tex]The graph of the function with its asymptotes can be seen below.
Erik puts $6,000.00 into an account to use for school expenses. The account earns 5%interest, compounded monthly. How much will be in the account after 8 years?
Answer:
[tex]A=\text{ \$8,944 after 8 years.}[/tex]Step-by-step explanation:
Compounded interest is represented by the following equation:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \text{where,} \\ P=\text{ principal invested} \\ r=\text{ interest rate} \\ n=\text{ number of times compounded per unit t} \\ t=\text{ time in years} \end{gathered}[/tex]Then, for a principal of $6,000.00 at a 5% interest, compounded monthly. After 8 years Erik would have:
[tex]\begin{gathered} A=6000(1+\frac{0.05}{12})^{96} \\ A=\text{ \$8,943.51} \\ \text{ Rounding to the nearest cent:} \\ A=\text{ \$8,944 after 8 years.} \end{gathered}[/tex]Quadrilateral TUWX has vertices T(-2, 1), U(4,2), and W(1,5), X(-1,-2), then itwas translated 2 units down. After the translation, what would be the newordered pair for point X?)*X(-1,-4)X(1,3)X(4,0)X(-2,-1)
What is the area of the figure?Assume that all angles are right angles A) 10.64 cm^2B) 12.54 cm^2C) 15.10 cm^2D) 16.24 cm^2
Area of the figure = 12.54 cm² (option B)
Explanation:To solve the question, we will divide the figure into shapes we can find their areas:
Area of A = Area of rectangle
Area of rectangle = length × width
length = 1.9cm, width = 1cm
Area of A = 1.9 × 1
Area of A = 1.9 cm²
Area of rectangle B = Area of rectangle
length = 5.6cm, width = 1.9cm
Area of B = 5.6 × 1.9
Area of B = 10.64 cm²
Area of the figure = Area of A + Area of B
Area of the figure = 1.9 + 10.64
Area of the figure = 12.54 cm² (option B)
Find percent notation for2/10
Find the indicated probability. Express your answer as a simplified fraction unless otherwise noted.
This is a conditional probability problem.
Let A be the event that the person is over 40
Let B be the event that the person is a root beer drinker
Probability of A given B is given by the equation:
[tex]P(A\text{/B)=}\frac{P(AnB)}{P(B)}[/tex][tex]\begin{gathered} P(\text{AnB)}=\frac{AnB}{Total\text{ number of subjects}} \\ P(\text{AnB)}=\frac{30}{255} \\ P(B)=\frac{number\text{ of root b}eer\text{ drinker}}{\text{Total number of subjects}} \\ P(B)=\frac{25+20+30}{255} \\ P(B)=\frac{75}{255} \end{gathered}[/tex]Thus,
[tex]\begin{gathered} P(A\text{/B)=}\frac{30}{255}\div\frac{75}{255} \\ P(A\text{/B)=}\frac{30}{255}\times\frac{255}{75} \\ P(A\text{/B)=}\frac{30}{75} \\ P(A\text{/B)=}\frac{2}{5} \end{gathered}[/tex]Hence , the correct option is option B
A seed company planted a floral mosaic of a national flag. The perimeter of the flag is 2,100 Determine the flag's width and length if the length is 410 ft greater than the width.
The perimeter of the flag, which is a rectangle, is equal to twice the length plus twice the width:
[tex]P=2l+2w[/tex]Let "w" represent the width of the rectangle. If the length is 410ft greater than the width, you can express the length as follows:
[tex]l=w+410[/tex]Replace the expression above in the formula of the perimeter:
[tex]P=2(w+410)+2w[/tex]You know that the perimeter is P= 2,100ft, then:
[tex]2100=2(w+410)+2w[/tex]The next step is to solve for w:
- Distribute the multiplication on the parentheses term:
[tex]\begin{gathered} 2100=2\cdot w+2\cdot410+2w \\ 2100=2w+820+2w \\ \end{gathered}[/tex]- Order the like terms together and simplify:
[tex]\begin{gathered} 2100=2w+2w+820 \\ 2100=4w+820 \end{gathered}[/tex]- Subtract 820 to both sides of the equal sign
[tex]\begin{gathered} 2100-820=4w+820-820 \\ 1280=4w \end{gathered}[/tex]- Divide both sides by 4
[tex]\begin{gathered} \frac{1280}{4}=\frac{4w}{4} \\ 320=w \end{gathered}[/tex]Now that the width is determined, you can calculate the length of the flag:
[tex]\begin{gathered} l=w+410 \\ l=320+410 \\ l=730 \end{gathered}[/tex]The width of the flag is 320ft and the length is 730ft
A shipping container will be used to transport several 50-kilogram crates across the country by rail. The greatest weight that can be loaded into the container is 24000 kilograms. Other shipments weighing 11600 kilograms have already been loaded into the container. Write and solve an inequality which can be used to determine xx, the number of 50-kilogram crates that can be loaded into the shipping container.
The greatest weight that can be loaded into the container is 24000 kilograms.
Let y be the weight of the container
[tex]y\le24000[/tex]Already loaded 11600 Kg in the container.
[tex]y\le24000-11600[/tex][tex]y\le12400[/tex]It can be loaded with several 50kg crates.
[tex]50x\le y[/tex]The equation to find the x number of 50 kg crates is
[tex]50x\le12400[/tex][tex]x\le\frac{12400}{50}[/tex][tex]x\le248[/tex]Hence the number of 50-
how do I rewrite 2x^2 + 2y^2 -8x +10y +2=0in standard form. this is the standard form for the equasion of a circle.
Student could not send the picture of the exercise and he is working on that.
Session ended by tutot.
Use completing the square to identify the solutions to the quadratic below: d^2 + 4d - 10 = 0
ANSWER
[tex]\begin{gathered} d=-2+\sqrt[]{14} \\ d=-2-\sqrt[]{14} \end{gathered}[/tex]EXPLANATION
We want to solve the quadratic equation using completing the square method:
[tex]d^2+4d-10=0[/tex]The general form of a quadratic equation is:
[tex]ax^2+bx+c=0[/tex]where a, b and c are coefficients.
The first step is to find a number that is equal to:
[tex](\frac{b}{2})^2[/tex]From the given equation, b is 4.
So, we have that:
[tex](\frac{4}{2})^2=2^2=4[/tex]Now, we can add that number to both sides of the equation and write it in this form:
[tex](d^2+4d+4)-10=4[/tex]Factorize the part of the left hand side in the parantheses:
[tex]\begin{gathered} (d^2+2d+2d+4)-10=4 \\ (d+2)^2-10=4 \end{gathered}[/tex]Add 10 to both sides of the equation:
[tex]\begin{gathered} (d+2)^2=4+10 \\ (d+2)^2=14 \end{gathered}[/tex]Find the square root of both sides:
[tex]\begin{gathered} d+2=+\sqrt[]{14} \\ \text{and} \\ d+2=-\sqrt[]{14} \end{gathered}[/tex]Subtract 2 from both sides of the equation:
[tex]\begin{gathered} \Rightarrow d=-2+\sqrt[]{14} \\ \text{and} \\ \Rightarrow d=-2-\sqrt[]{14} \end{gathered}[/tex]That is the solution of the equation according to completing the square method.
The table shows the elevations of the highest and lowest points in France, where negative numbers are feet below sea level. Which inequality correctly compares the elevations? Point Elevation (ft) Mont Blanc 15,771 Rhone River delta-7 O A. 15,771 <-7O B. 15,771 >-7O C. -7 > 15,771O D. 7 > 15,771
We have to compare the numbers -7 and 15, 771.
As the Rhone River delta is found in a lower point compared to Mont Blanc, this means that the number (-7) should be less than 15,771. This is:
[tex]-7<15,771[/tex]This is equivalent to say that the number 15,771 is greater than -7, or:
[tex]15,771>-7[/tex]Thus, the inequality that correctly compares the elevataions is 15,771 > -7.
Suppose that the polynomial function f is defined as follows.f(x)=3x³(x-4)²(x - 5)(x+7)³List each zero off according to its multiplicity in the categories below.If there is more than one answer for a multiplicity, separate them with commas. If there is no answer, click on "None."Zero(s) of multiplicity one:Zero(s) of multiplicity two:Zero(s) of multiplicity three:08 0.0....XNoneŚ
f(x)=3x³(x-4)²(x - 5)(x+7)³
To find the zeros, set the function equal to zero
3x³(x-4)²(x - 5)(x+7)³ =0
Using the zero product property
3x^3 =0 ( x-4) ^2 =0 x-5 =0 ( x+7) ^3 =0
Solving each equation
x=0 x=4 x=5 x = -7
We can find the multiplicity from the power of the factor
x=0 has a multiplicity of 3
x =4 has a multiplicity of 2
x =5 has a multiplicity of 1
x = -7 has a multiplicity of 3
Zero(s) of multiplicity one:5
Zero(s) of multiplicity two:4
Zero(s) of multiplicity three: 0,-7
Complete the description of the procedure performed to derive the slope-intercept form of a linear equation. First let m be the slope and b be the y-Intercept of an arbitrary line, L. Because b is the y-Intercept, (select)y must be on the line. change in y y2-y1 Substitute (0, b) into (select) change in x X2 - X1 Finally you simplify this equation to arrive at (select)
Given data:
The given y-intercept is b.
The given slope is m.
The expression for the equation of the line is,
[tex]y=mx+b[/tex]Thus, the equation of the line is y=mx+b.
