In the diagram below, lines k and intersect lines
m and n at points A, B, C, and D
Which statement is sufficient to prove ABCD is a
parallelogram?
1) Angles 1 and 3
2)angles 4 and 7
3)angles 2 and 5, angles 5 and 7
4) angles 1 and 3, angles 3 and 4
Answers
Answer:
actually c is the right answer
Angles 2 = angle 5, and angles 5 = angle 7. Option 3 is correct.
What is quadrilateral?A quadrilateral is a four-sided shape as given in the picture in question.
What is the angle?Orientation of the one line with respect to the horizontal or other respective line is known as a measure of orientation and this measure is known as the angle.
here,
In the diagram,
Lines k and intersect lines m and n at points A, B, C, and D,
So,
For ABCD, proved to be a parallel gram, ,
Angles 2 = angle 5, and angles 5 = angle 7
Thus, Angles 2 = angle 5, and angles 5 = angle 7. Option 3 is correct.
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Related Questions
5. The function is continuous on the interval [10, 20] with some of its values given in the table above. Estimate the average value of the function with a Right Hand Sum Approximation, using the intervals between those given points. (4 points)x1012151920f(x)-2-5-9-12-16A -9.250B -10.100C-7.550D-6.700
Answers
Given the following question:
P(x)=(x+5)(x-5) and q(x)=(x+3)(x-3) where do the graphs of the two intersects
Answers
For the equation P(x)=(x+5)(x-5) and q(x)=(x+3)(x-3) the graph of the two equation never intersect each other.
What is a conic section?It is defined as the curve which is the intersection of cone and plane. There are three major conic sections; parabola, hyperbola, and ellipse (a circle is a special type of ellipse).
The given equations are,
P(x)=(x+5)(x-5)
q(x)=(x+3)(x-3)
Parabola is defined as the graph of a quadratic function that has something bowl-shaped.
(x - h)² = 4a(y - k)
(h, k) is the vertex of the parabola:
a = √[(c-h)² + (d-k²]
(c, d) is the focus of the parabola:
The given equation represents a parabola as the equation is graphed we see that the graph of the two never intersects each other.
Thus,for the equation P(x)=(x+5)(x-5) and q(x)=(x+3)(x-3) the graph of the two equation never intersect each other.
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15.4-32+(60/3*166)*8 divided by 4-(2*61)
Answers
The 15.4-32+(60/3*166)*8 divided by 4-(2*61) is -224.94.
As per the PEMDAS rule, firstly solving the parenthesis in the numeral : 15.4-32+(60/3*166)*8
Performing division in parenthesis
Number = 15.4 - 32 + (20×166) × 8
Performing multiplication in parenthesis
Number = 15.4 - 32 + 3320 × 8
Performing multiplication and subtraction
Number = - 16.6 + 26,560
Performing subtraction
Number = 26,543.4
Number = 4 - (2×61)
Performing multiplication in parenthesis
Number = 4 - 122
Performing subtraction
Number = - 118
Performing division now
Result = 26,543.4 ÷ -118
Result = -224.94
The number obtained on division will be -224.94.
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Ten light bulbs were in a chandelier. Three-fifths of the bulbs were shining. What fraction of the light bulbs were not shining?
Answers
2/5 of bulbs were not shining. It is equal to 4 bulbs.
Step-by-step explanation:
10/10 bulbs equals all 10 bulbs. 3/5 were shining, that is equal to 6/10 of all ten bulbs. (or 60%). 60% of ten is 6 bulbs shining.The number of bulbs that werent shining is 10 - 6=4.
This is algebra 2 ( function graphs) I’m usually okay with math but I been I of school recently for surgery and forgot a little bit I just need a refresher
Answers
The red function has the form of:
[tex]f(x)=\sqrt[]{-x+1}[/tex]In order to obtain the green function, we need to do a reflection over y-axis, then a translation one unit to the right, a translation 1 unit down, and finally a reflection over the x-axis, so:
[tex]\begin{gathered} g(x)=-(\sqrt[]{-(-x)+1-1}-1) \\ g(x)=-\sqrt[]{x}+1 \end{gathered}[/tex]7. The positive interval (s) of the functiony=-x ²1 are
Answers
Let's graph the function and find the positive intervals, as follows:
Therefore, the interval is : (-∞ , +-∞)
write each of the following numbers line position as fraction with Demeter 100 as decimals and also as percentages
Answers
Answer:
Explanation:
To write the given numbers as fractions of 100, percentages, and decimals, we first need to estimate their values on the number line. Once, we have the values of the numbers, we can write the as a fraction of 100 as
[tex]\frac{Num}{100}[/tex]As percentages as
[tex]\frac{Num}{100}\times100[/tex]And as decimals as
[tex]Num\div100[/tex](a).
The estimate of the value of three numbers is 27, 45, 67.
Writing the above as fractions of 100 gives
[tex]\frac{27}{100},\frac{45}{100},\frac{67}{100}[/tex]As a percentage, these numbers are
[tex]\frac{27}{100}\times100,\frac{45}{100}\times100,\frac{67}{100}\times100[/tex][tex]\rightarrow27\%,45\%,67\%[/tex]To write the numbers as decimals we divide them by 100 to get
[tex]0.27,0.45,0.67[/tex](remember that dividing by 100 shifts the decimal point to the left by 2 digits)
(b).
The estimate of the values of the three numbers are 57, 74, and 89
Writing these numbers as fractions gives
[tex]\frac{57}{100},\frac{74}{100},\frac{89}{100}[/tex]As a percentage these numbers are
[tex]\frac{57}{100}\times100,\frac{74}{100}\times100,\frac{89}{100}\times100[/tex][tex]57\%,74\%,89\%[/tex]And as decimals
[tex]0.57,0.74,0.89[/tex](c).
The estimate of the value of the three numbers is 22, 36, 55.
Writing them as a fraction gives
[tex]\frac{22}{100},\frac{36}{100},\frac{55}{100}[/tex]As a per cent these numbers are written as
[tex]undefined[/tex]Fitness Works charges a $20 monthly fee, plus $5 for each class you take. Gym-tastic charges$100 monthly fee, and offers FREE unlimited classes. How many classes do you have to take forthe cost to be the exact same at both gyms?
Answers
Let us assume that the number of classes is x and the total fee is y
In the Fitness works, there is a monthly fee of $20 and $5 per class, then
The total fee y = 20 + 5(x), then
y = 20 + 5x ------ (1)
In the Gym-tastic, there is a monthly fee of $100 for unlimited classes, then
y = 100 ------- (2)
Equate (1) and (2)
20 + 5x = 100
Subtract 20 from both sides
20 - 20 + 5x = 100 - 20
5x = 80
Divide both sides by 5 to find x
x = 16
The number of the classes is 16
The radius of a cylinder is 8cm. it's height is three times it's radius. What is the surface area of the cylinder? No pictures available
Answers
Recall that the surface area of the cylinder is
[tex]\begin{gathered} SA=2\pi rh+2\pi r^2 \\ \text{where} \\ r\text{ is the radius} \\ h\text{ is the height} \end{gathered}[/tex]Given the following
radius of 8 cm, and height of 24 cm (3 times the radius), then the surface area of the cylinder is
[tex]\begin{gathered} SA=2\pi rh+2\pi r^2 \\ SA=2\pi(8\text{ cm})(24\text{ cm})+2\pi(8\text{ cm})^2 \\ SA=384\pi\text{ cm}^2+128\pi\text{ cm}^2 \\ SA=512\pi\text{ cm}^2 \\ \; \\ \text{Therefore, the surface area of the cylinder is }512\pi\text{ cm}^2 \end{gathered}[/tex]Which of these is a point-slope equation of the line that is perpendicular toy-25 = 2(x-10) and passes through (-3,7)?-O A. y+ 7 = 2(x-3)O B. y- 7 = -2(x+3)O C. y-7=-(x+3)O D.y+7=-1(x-3)-
Answers
We have to find the equation of the line, in point-slope form, that is perpendicular to y - 25 = 2(x - 10) and passes through (-3,7).
The line y - 25 = 2(x - 10) has a slope m = 2.
Perpendicular lines have slopes that are negative reciprocals, so our line will have a slope that is:
[tex]m=-\frac{1}{m_p}=-\frac{1}{2}[/tex]Then, we have the slope m = -1/2 and the point (-3,7), so we can write the point-slope form of the equation as:
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-7=-\frac{1}{2}(x-(-3)) \\ y-7=-\frac{1}{2}(x+3) \end{gathered}[/tex]Answer: y - 7 = -1/2 * (x + 3) [Option C]
what products of 37 and 4
Answers
A product is the result of a multiplication between 2 numbers:
37 x 4 = 136
Which sentence is true about an equilateral triangle?
Answers
In Equilateral triangle all sides are equal.
8. T : R 2 ---+ R 2 first reflects points through the horizontal x 1-
axis and then reflects points through the line x2 = x 1•
Answers
Answer:
Thats nice
Step-by-step explanation:
Select all the correct answers.Which vectors are unit vectors?1 一32l -一》《完,表》口 u={1, 1}1-(美)
Answers
The unit vector has a magnitude = 1
So, for the given vectors, we will find the magnitude of every vector
[tex]\begin{gathered} u=<\frac{\sqrt[]{3}}{2},-\frac{1}{2}> \\ |u|=\sqrt[]{(\frac{\sqrt[]{3}}{2})^2+(-\frac{1}{2})^2}=\sqrt[]{\frac{3}{4}+\frac{1}{4}}=\sqrt[]{\frac{4}{4}}=\sqrt[]{1}=1 \end{gathered}[/tex]So, it is a unit vector
[tex]\begin{gathered} u=<-\frac{2}{\sqrt[]{5}},\frac{1}{\sqrt[]{5}}> \\ |u|=\sqrt[]{(-\frac{2}{\sqrt[]{5}})^2+(\frac{1}{\sqrt[]{5}})^2}=\sqrt[]{\frac{4}{5}+\frac{1}{5}}=\sqrt[]{\frac{5}{5}}=1 \end{gathered}[/tex]So, it is a unit vector
[tex]\begin{gathered} u=<1,1> \\ |u|=\sqrt[]{1^2+1^2}=\sqrt[]{1+1}=\sqrt[]{2}=1.414 \end{gathered}[/tex]So, it is not a unit vector
[tex]\begin{gathered} u=<-\frac{5}{\sqrt[]{6}},\frac{1}{\sqrt[]{6}}> \\ |u|=\sqrt[]{(-\frac{5}{\sqrt[]{6}})^2+(\frac{1}{\sqrt[]{6}})^2}=\sqrt[]{\frac{25}{6}+\frac{1}{6}}=\sqrt[]{\frac{26}{6}}=2.08 \end{gathered}[/tex]So, it is not a unit vector
So, the correct options are: 1 and 2
Create a table to show the relationship of values of X and values of y
Answers
We have to complete a table with some points (x,y) from the line that is represented in the graph.
To do that we choose a value of x and wee which value of y corresponds to that value of x in the line.
For example, we can do it for x = 1 as:
Then, we have one point for the table: when x = 1, y = -9.
We can repeat this process for some points of x:
x | y
-------------
-4 | 1
-3 | -1
-2 | -3
-1 | -5
0 | -7
1 | -9
2 | -11
We can see that for each unit increase in x, the value of y decreases by 2. This indicates that the slope is m = -2.
Also, for x = 0, y = -7. Then b = -7 is the y-intercept.
which of the following values is the solution to the equation -16 + x equals 30
Answers
If we have the equation:
[tex]-16+x=30[/tex]We can solve it as follows:
1. Add 16 to both sides of the equation (addition property of equality):
[tex]-16+16+x=30+16[/tex]Then, we have:
[tex]x=30+16\Rightarrow x=46[/tex]Therefore, the value for x is equal to 46 (x = 46). We can check this as follows:
[tex]-16+46=30\Rightarrow30=30[/tex]We substituted the value, x = 46, in the original equation. This is always TRUE. Therefore, the value for x = 46.
If the GCF of the nurmerator and the denominator is 1, then the fraction is in __
Answers
Recall that a fraction is of the form
[tex]\frac{a}{b}[/tex]where a is the numerator and b is the denominator. The GCF of two numbers is the biggest number that is less than both numbers and that it divides them without any remainder. When the GCF of the numerator and the denominator is 1. This means that we cannot find a common factor for both numbers, so we cannot cancel any more factors. This leads to the fact that the fraction is irreducible or that it is in its simplest form.
I need help with this I have only questions1. Where is y when x is 02. find f(-4)3. what is x when y is 4?4. What is X when f(X)=0? There are two answers put the smaller number in the first answer blank ____ and ____
Answers
1) y = 1 2) y =3 3) x = -3 4) y= -7 and y = 1
We are to answer the questions stated using the graph:
1) From the graph, when x = 0:
To understand this, check the point on the line that only lies on y
y = 1
2) f(-4) is the same as what is the value of y when x = -4
This is because f(x) = y
From the graph, when x = -4
y = 3
3) From the graph, when y =4
Trace the value of y=4 on the line to get corresponding x value:
x= -3
4) When f(x) = 0
This means what is the value of x when y = 0. This is because f(x) = y.
From the graph, we have two values of x when y = 0
y = -7
y = 1
To fill the blank spaces starting with the smaller number:
-7 and 1
255 is 88% of what? Round to the nearest tenth.
Answers
Answer:
289.8.
Explanation:
Let the number = x
Then, it implies that:
[tex]88\%\text{ of }x=255[/tex]Next, solve for x:
[tex]\begin{gathered} \frac{88}{100}\times x=255 \\ 88x=25500 \\ x=\frac{25500}{88} \\ x=289.8 \end{gathered}[/tex]The number is 289.8 to the nearest tenth.
I have no idea how to solve this I have to find the missing terms outside the box
Answers
Notice that:
[tex]\begin{gathered} 18x^3=3x\times6x^2, \\ -3x^2=3x\times(-x), \\ 27x=3x\times9. \end{gathered}[/tex]Answer:
Question 5 of 6Which exponential expression is equivalent to the one below?(22• (-7))40A. 40 • (22• (-7))O B. (22) • (-7)40C. (22)40 + (-7) 40D. (22)40 . (-7)40+SUBMIT
Answers
Okay, here we have this:
Considering the provided expression, we are going to analize which exponential expression is equivalent, so we obtain the following:
As the property of the exponent of a multiplication says that it is equal to the product of each number raised to that power. We have this:
[tex]\begin{gathered} \mleft(22\cdot\mleft(-7\mright)\mright)^{40} \\ =(22)^{40}\cdot(-7)^{40} \end{gathered}[/tex]Finally we obtain that the correct answer is the option D.
30. Figure A has an area of 18 sq. ft. Figure B has anarea of 98 sq. ft. and one side length is 14 ft. What isthe corresponding side length of Figure A?
Answers
Remember that
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
In this problem
Figure A and Figure B are similar
so
step 1
Find out the scale factor
scale factor^2=18/98
scale factor=√(18/98)
step 2
To find out the corresponding side length of Figure A, multiply the side length of figure B by the scale factor
so
14*√(18/98)=6
the answer is 6 ftwhat is the solution for 7+k<35
Answers
We are given the following inequation:
[tex]7+k<35[/tex]To find the solution we need to subtract 7 to both sides, like this:
[tex]\begin{gathered} 7-7+k<35-7 \\ k<28 \end{gathered}[/tex]Therefore, the solution is the values of "k" smalled than 28.
What is the quotient of the complex numbers below?A.10 - iB.2 - iC.2 + iD.10 + i
Answers
The given fraction is
[tex]\frac{7+i}{3-i}[/tex]We will multiply up and down by the conjugate of down (3 + i)
[tex]\begin{gathered} \frac{7+i}{3-i}\times\frac{3+i}{3+i}= \\ \\ \frac{(7+i)(3+i)}{(3-i)(3+i)}= \\ \\ \frac{(7)(3)+(7)(i)+(i)(3)+(i)(i)}{(3)(3)-(i)(i)}= \end{gathered}[/tex]Add the like terms
[tex]\begin{gathered} \frac{21+7i+3i+i^2}{9-i^2}= \\ \\ \frac{21+10i-1}{9-(-1)}= \\ \\ \frac{20+10i}{9+1}= \\ \\ \frac{20+10i}{10} \end{gathered}[/tex]Take 10 as a common factor from up
[tex]\begin{gathered} \frac{10(2+i)}{10}= \\ \\ 2+i \end{gathered}[/tex]The answer is C
Kathy and Cheryl are walking in a fundraiser. Kathy completes the course in 3.6 hours and Cheryl completes the course in 6 hours. Kathy walks two miles per hour faster than Cheryl. Find Kathy's speed and Cheryl's speed in miles per hour.
Answers
Answer:
Explanation:
Here is what we know;
Both Kathy and Cheryl cover the same distance ( one course).
Kathy completes the course in 3.6 hours.
Kathy's speed is 2 miles/hour faster than Cheryl's
Cheryl completes the course in 6 hours
Cheryl's speed is yet unkown.
Now, the speed v is defined as
[tex]v=\frac{D}{t}[/tex]where D is the distance covered and t is the time taken.
Now, let us say D = distance of one course. Then in Kathy's case, we have
[tex]v_{\text{kathy}}=\frac{D}{3.6hr}[/tex]since Kathy's speed is 2 miles per hour faster than Cheryl's, we have
[tex]v_{\text{kathy}}=\frac{D}{3.6}=2+v_{\text{cheryl}}[/tex]For Cheryl, we know that
[tex]v_{\text{cheryl}}=\frac{D}{6hr}[/tex]or simply
[tex]v_{\text{cheryl}}=\frac{D}{6}[/tex]Putting this into the equation for Kathy's speed gives
[tex]v_{\text{kathy}}=\frac{D}{3.6}=2+v_{\text{cheryl}}\Rightarrow v_{\text{kathy}}=\frac{D}{3.6}=2+\frac{D}{6}[/tex][tex]\Rightarrow\frac{D}{3.6}=2+\frac{D}{6}[/tex]We have to solve for D, the distance of a course.
Subtracting D/6 from both sides gives
[tex]\frac{D}{3.6}-\frac{D}{6}=2[/tex][tex](\frac{1}{3.6}-\frac{1}{6})D=2[/tex][tex]\frac{1}{9}D=2[/tex][tex]D=18\text{miles}[/tex]Hence, the distance of a course is 18 miles.
With the value of D in hand, we can now find the velocity of Kathy and Cheryl.
[tex]v_{\text{cheryl}}=\frac{D}{6hr}=\frac{18\text{miles}}{6hr}[/tex][tex]\Rightarrow\boxed{v_{\text{cheryl}}=3\text{ miles/hr}}[/tex]Hence, Cheryl's speed is 3 miles/hr.
Next, we find Kathy's speed.
[tex]v_{\text{kathy}}=\frac{D}{3.6hr}=\frac{18\text{miles}}{3.6hr}[/tex][tex]\boxed{v_{\text{kathy}}=5\text{miles}/hr\text{.}}[/tex]Hence, Kathy's speed is 5 miles/hr.
Therefore, to summerise,
Kathy's speed = 5 miles/hr
Cheryl's speed = 3 miles/hr
In standard notation, 208.06 is written as:
Answers
Answer:
[tex]2.0806 * 10^{2}[/tex]
Step-by-step explanation:
Standard notation, also known as scientific notation, is when the whole number can only be [tex]1\leq 0 < 10[/tex]. The rest of the number is behind the decimal point.
hello so which equivalent to [tex]30 \div (3 + )[/tex]
Answers
Answer:
G. 30 ÷ (x+3)
Explanation:
Given the expression:
[tex]30\div(3+x)[/tex]From the given options, note that:
[tex]3+x=x+3[/tex]Therefore, an equivalent expression is:
[tex]30\div(x+3)[/tex]The correct choice is G.
HELP PLEASE THIS IS DUE. I been asking for a while but I just get spam answers.
Answers
Answer:
y = 12
Step-by-step explanation:
Consider the triangles in the diagram. Triangle QRS (the smaller one on the left) and Triangle PRO (the whole shape)
These two triangles are similar. It helps to write them separately. See image.
You can use a proportion (two ratios equal to each other) to solve this.
There are two good ways to set up an equation.
EITHER:
bottomLeg/sideLeg=bottomLeg/sideLeg
OR
smallbottom/bigbottom=smallside/bigside
see image.
Either way you set it up the answer comes out the same. Pretty much all the work is the same after you crossmultiply.
Solve 9/y = 12/16
OR 9/12 = y/16
see image.
Use square roots for the problem. Which equation(s) have -4 and 4 as solutions? Select all that apply
Answers
Given:
Different equations
To find:
the equation whose solutions have -4 and 4
To determine the equations with solutions -4 and 4, we will solve each of th given equation
[tex]\begin{gathered} a)\text{ x}^2\text{ = 8} \\ x\text{ = }\pm\sqrt{8}\text{ = }\pm\sqrt{4\times2} \\ x\text{ = }\pm\text{2}\sqrt{2}\text{ \lparen not a solution of -4 and 4\rparen} \end{gathered}[/tex][tex]\begin{gathered} b)\text{ x}^2\text{ + 16 = 0} \\ x^2=\text{ -16} \\ x\text{ = }\pm\sqrt{-16} \\ root\text{ of -16 gives a complex number. Hence, no solution of -4 and 4} \end{gathered}[/tex][tex]\begin{gathered} c)\text{ 2x}^2\text{ = 32} \\ divide\text{ both sides by 2:} \\ x^2\text{ = }\frac{32}{2} \\ x^2\text{ = 16} \\ x\text{ = }\pm\sqrt{16}\text{ = }\pm4 \\ x\text{ = 4 and -4} \end{gathered}[/tex][tex]\begin{gathered} d)\text{ -3x}^2\text{ = -48} \\ divide\text{ both sides by -1:} \\ division\text{ of same signs give positive sign} \\ 3x^2\text{ = 48} \\ x^2\text{ = 48/3} \\ x^2\text{ = 16} \\ x\text{ = }\pm\sqrt{16}\text{ = }\pm4 \\ x\text{ = 4 and - 4} \end{gathered}[/tex][tex]\begin{gathered} e)\text{ }6x^2\text{ + 56 = -40} \\ 6x^2\text{ = -40 - 56} \\ 6x^2\text{ = -96} \\ x^2\text{ = -96/6} \\ x^2\text{ = -16} \\ x\text{ = }\pm\sqrt{-16} \\ root\text{ of a negative number gives a complex number.} \\ Hence,\text{ no solution of -4 and 4} \end{gathered}[/tex][tex]\begin{gathered} f)\text{ 27 - 5x}^2\text{ = -53} \\ add\text{ 5x}^2\text{ }to\text{ both sides:} \\ 27\text{ - 5x}^2+\text{ 5x}^2\text{ = -53 + 5x}^2 \\ 27\text{ = -53 + 5x}^2 \\ \\ add\text{ 53 to btoh sides:} \\ 27\text{ + 53 = 5x}^2 \\ 80\text{ = 5x}^2 \\ divide\text{ both sides by 5:} \\ \frac{80}{5}=\text{ x}^2 \\ x^2\text{ = 16} \\ x\text{ = }\pm\sqrt{16}\text{ = }\pm4 \\ x\text{ = 4 and -4} \end{gathered}[/tex]let E b the event where the sun of two rolled dice is greater than or equal to 3. lost the outcomes in E^c
Answers
We have the event E defined as:
E: The sum of two rolled dice is greater than or equal to 3
The event E^c is the negation of the event E. Then:
E^c: The sum of two rolled dice is smaller than 3
For two dice, the minimum sum is 2, so this is equal to the event "the sum of two rolled dice is 2". There is only one outcome:
Dice 1: 1
Dice 2: 1
Sum: 2
Then, the only possible outcome for E^c is {1, 1}
