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3 years ago

Line QR contains (2, 8) and (3, 10) Line ST contains points (0, 6) and (-2, 2). Lines QR and ST are (4 points)

1 answer:

3 0

Slope formula:
m = y2 - y1 / x2 - x1

Line QR: (2,8) (3,10)
m = 10 - 8 / 3 - 2 = 2 / 1 = 2

Line ST: (0,6) (-2,2)
m = 2 - 6 / -2 - 0 = -4/-2 = 2

c. parallel because the slopes are the same

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Solve for h.

Luden [163]

Solving for h means we write the equation in such a way that h is isolated from other variables. It is done as follows:

(P = mgh) 1/mg
P/mg = mgh/mg
h = P/mg

Therefore, the correct answer from the choices listed above is option C, h = p/mg.

Hope this answers the question. Have a nice day.

4 0

3 years ago

Solve for x. I was sure the answer was 2? Please show me the breakdown. THX

Murljashka [212]

Answer:

x = 8 and y = 10

Step-by-step explanation:

Since the figures are congruent:

Angle PQR = EFG

6y + x = 68

Segment QR = FG

2x - 4 = 12

Solving that:

2x - 4 = 12

2x = 16

x = 8

Replacing x in 6y + x =68

6y + 8 = 68

6y = 60

y = 10

7 0

4 years ago

What is the product of (9)(-9)(-1)?

LekaFEV [45]

The product of this is 81

5 0

3 years ago

2 - X<-7 (x > 9) solve and graph

NARA [144]

Use Photomath it graphs too

5 0

4 years ago

Give the equation for a circle with the given center and radius.Center at (-10, -4), radius = 5A. (x−10)^2+(y−4)^2=25B. (x−4)^2+

BaLLatris [955]

$D)(x+10)^2+(y+4)^2=5^2=25$

Explanation

Step 1

We know that the general equation for a circle is

$\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ \end{gathered}$

where

(h,k) is the center of the circel

and r is the radius

then, all we need to do is replace

Let

$\begin{gathered} \text{center}\Rightarrow(h,k)\Rightarrow(-10,-4) \\ \text{radius}\Rightarrow r\Rightarrow5 \end{gathered}$

replace the values

$\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (x-(-10))^2+(y-(-4))^2=5^2 \\ (x+10)^2+(y+4)^2=5^2=25 \end{gathered}$

therefore, the answer is

$D)(x+10)^2+(y+4)^2=5^2=25$

I hope this helps you

7 0

2 years ago