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schepotkina [342]
2 years ago
9

Help me with this exercise

Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
3 0

$\left(\frac{-3}4\right)^3 = \frac{-3}4 \times \frac{-3}4 \times \frac{-3}4 = \boxed{-\frac{27}{64}}$

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Ramona is counting the posts between mile markers on the highway. In 1 mile, she counts 33 posts. If the posts are evenly spaced
Naddika [18.5K]

Answer:

160 feet

Step-by-step explanation:

In order to solve this, we need to realize that the mile is being divided into 33 equal parts. Therefore our operation should look like this:

5,280 (the number of feet in 1 mile) / 33

5,280/33=160

The posts are 160 feet apart from each other.

HTH :)

5 0
2 years ago
Simplify the expression 6h+​(−8.7d​)−13+6d−3.3h.<br> 6h+​(−8.7d​)−13+6d−3.3h=
mezya [45]

Answer:

 -2.7d+2.7h-13

6 0
3 years ago
I don’t get how to get the answer or graph it
timofeeve [1]
The distance might be the slope of the line
7 0
2 years ago
What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?
xxMikexx [17]

We have been given an equation of hyperbola \frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1. We are asked to find the center of hyperbola.  

We know that standard equation of a vertical hyperbola is in form \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1, where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1

Now we can see that value of k is -3. Therefore, the vertex of given hyperbola will be at point (6,-3) and option D is the correct choice.

6 0
3 years ago
PROBABILITY. Areas of regions: Entire Target=84in^2, Red area=28in^2, Green area=7in^2, Yellow area=9.42in^2, Blue area=39.58in^
abruzzese [7]

Answer:

HIIIIIIIIIIIIIIIIIIIIIIIIII

Step-by-step explanation:

8 0
3 years ago
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