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2 years ago

Please help im timed with 3 mins left! will mark branliest

Mathematics

2 answers:

Serjik [45]2 years ago

7 0

It’s D hope this helps with your assignment

ad-work [718]2 years ago

3 0

Answer: ITS D

Step-by-step explanation:

Imma just answer real quikc since its timed if u want explanation just ask

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Wendy is on a long road trip. So far, she has driven 125 miles in 150 minutes.

Vikentia [17]

Answer:

Step-by-step explanation:

125 miles / 150 minutes = 300 miles / x Criss multiply

Notice that the numerators both have the same units -- miles. Proportions should always be set up that way.

125 x = 150 * 300 Combine the right

125x = 45000 Divide by 125

x = 45000/125

x = 360 more minutes which is 6 hours.

6 0

2 years ago

6. How could you change the simulation to generate data for three pets? Result Frequency Heads, Heads 9 Heads, Tails 14 Tails, H

ArbitrLikvidat [17]

9 plus 14 plus 18 plus 9 is 50

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2 years ago

Find the product: (-4)(-8)

mario62 [17]

Answer: The product of (-4)(-8) is 32. Multiply two negatives makes it positive. So just think of it as 4 times 8. Which is 32.

5 0

2 years ago

Read 2 more answers

Given: △ABC, AB=5 square root 2 <br><br>m∠A=45°, m∠C=30°<br><br>Find: BC and AC

hichkok12 [17]

Answer:

Part a) $BC=10\ units$

Part b) $AC=13.66\ units$

Step-by-step explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$C=30^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}$

solve for BC

$BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))$

$BC=10\ units$

step 2

Find the measure of angle B

we know that

The sum of the interior angles in a triangle must be equal to 180 degrees

$m\angle A+m\angle B+m\angle C=180^o$

substitute the given values

$45^o+m\angle B+30^o=180^o$

$75^o+m\angle B=180^o$

$m\angle B=180^o-75^o$

$m\angle B=105^o$

step 3

Find the length side AC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$B=105^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}$

solve for AC

$AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))$

$AC=13.66\ units$

4 0

2 years ago

Can someone help me?

Maslowich

Answer:

4. .88

5. 17/9

6. 67/12

7. 47/56

8. 209/45 or 4 29/45, 4.64

9. 143/30

Step-by-step explanation:

7 0

2 years ago