All categories

2 years ago

Given C(x, 16), D(2,-4), E(-6, 14), and F(-2, 4), find the value of x so that CD || EF.

Mathematics

1 answer:

Andrew [12]2 years ago

8 0

Answer:

x=-6

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are the same

In this problem

slope CD=slope EF

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope EF

we have

E(-6,14) and F(-2,4)

substitute the values in the formula

$m_E_F=\frac{4-14}{-2+6}\\m_E_F=\frac{-10}{4}\\m_E_F=-2.5$

step 2

Find the slope CD

we have

C(x, 16) and D(2, -4)

substitute the values in the formula

$m_C_D=\frac{-4-16}{2-x}\\m_C_D=\frac{-20}{2-x}$

Remember that

$m_C_D=m_E_F$

$\frac{-20}{2-x}=-2.5$

$-20=-5+2.5x\\2.5x=-20+5\\2.5x=-15\\x=-6$

You might be interested in

In triangle RST, U is the midpoint of RS, V is the midpoint of ST, and W is the midpoint of TR. Use the triangle diagram to answ

Zolol [24]

(1)

we are given

U is the midpoint of RS

and we have

$RU=12$

so, we can use formula

$RS=2RU$

we can plug values

$RS=2\times 12$

$RS=24$ ............Answer

(2)

V is the midpoint of ST

so, we get

$SV=VT$

now, we can plug values

$2x=11$

divide both sides by 2

$x=5.5$ .........Answer

(3)

now, we can find y

W is the midpoint of TR

so, we get

$RW=WT$

we can plug value

$15.9=3y$

divide both sides by 3

$y=5.3$

(4)

we can see that

triangles URW and RST are similar

so, their sides ratios must be equal

so, we get

$\frac{RW}{RT} =\frac{UW}{ST}$

we can plug values

$\frac{15.9}{2\times 15.9} =\frac{UW}{2VT}$

$\frac{15.9}{2\times 15.9} =\frac{UW}{2\times 11}$

$UW=22\times \frac{15.9}{2\times 15.9}$

$UW=11$ ...........Answer

(5)

we can see that

triangles SUV and RST are similar

so, their sides ratios must be equal

so, we get

$\frac{SU}{SR} =\frac{UV}{RT}$

now, we can plug values

$\frac{12}{2\times 12} =\frac{UV}{2\times 15.9}$

$UV=2\times 15.9\times \frac{12}{2\times 12}$

$UV=15.9$ .............Answer

4 0

2 years ago

Which statements about the hyperbola are true? Check all that apply.

mariarad [96]

Answer:

True options: 1, 2 and 5

Step-by-step explanation:

From the given diagram, you can see that the center of the hyperbola is placed at the origin, so first option is true (see attached diagram for definition of center, vertices, foci, i.e.)

There are two vertices of the hyperbola, they are placed at (-6,0) and (6,0), so second option is true.

The transverse axis is the segment connecting vertices, this segment is horizontal, so option 3 is false.

The foci are not placed within the rectangular reference box, so this option is false.

The directrices are vertical lines with equations $x=\pm \dfrac{a}{e}$ , so this option is true.

8 0

2 years ago

Read 2 more answers

A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.