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3 years ago

Find the midpoint of segment P Q if P(15, -3) and Q(5, 12)?

1 answer:

7 0

You can calculate this by averaging the X and Y coordinates independently:

X: (15+5)/2 = 10
Y: (-3+12)/2 = 4.5

Midpoint is at (10, 4.5)

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Given: m =-2/3

Elenna [48]

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The correct answer is (( D )) .

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6 0

2 years ago

4 (b - 3 ) = 6 steps in order from first to last

Amiraneli [1.4K]

Answer:

B= 9/2 Hope this helped <3

Step-by-step explanation:

Let's solve your equation step-by-step.

4(b−3)=6

Step 1: Simplify both sides of the equation.

4(b−3)=6

(4)(b)+(4)(−3)=6(Distribute)

4b+−12=6

4b−12=6

Step 2: Add 12 to both sides.

4b−12+12=6+12

4b=18

Step 3: Divide both sides by 4

8 0

3 years ago

Read 2 more answers

I need help with no solution

drek231 [11]

use photo math For that Problem It would work Mate.

4 0

3 years ago

Read 2 more answers

A bookstore sells textbooks for $80 each and notebooks for $4 dollars each. The bookstore would like to sell $800 in merchandise

mariarad [96]

Answer: 8 textbooks

Step-by-step explanation:

Sold 40 notebooks = 40 x $4 = $160

$800 - $160 = $640 left to sell

$640/$80 (cost of textbook) = 8 textbooks need to be sold

3 0

3 years ago

In triangle abc, m of acb = 90, cd is perpendicular to ab , m of acd is 60. and bd is 5 cm. find ad

weeeeeb [17]

Let us draw a picture to make the things more clear.

Attached is the image.

We have been given that

$\angle acd = 60 ^{\circ}$

Therefore, we have

$\angle dcb =90- 60= 30 ^{\circ}$

Now, in triangle bcd, we have

$\tan30 = \frac{5}{cd}\\
\\
\frac{1}{\sqrt 3}=\frac{5}{cd}\\
\\
cd=5\sqrt 3$

Now, in triangle acb, we have

$tan 60 = \frac{ad}{5\sqrt3} \\
\\
\sqrt 3= \frac{ad}{5\sqrt3}\\
\\
ad= 5\sqrt3 \times \sqrt 3\\
\\
ad= 5\times 3\\
\\
ad=15$

Thus, ad is 15 cm.

4 0

2 years ago