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jekas [21]
3 years ago
7

What is the approximate area of the shaded region? the shaded area is 28  A.168.56 square inches  B.696.08 square inches  C.740.

04 square inches  D.1677.76 square inches
Mathematics
1 answer:
svp [43]3 years ago
5 0
The answer is B its quite easy if you look at it
 
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Andy canoe is 2.7 meters long. how long would it be in milliters?
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A meter is 1000 millimeters.

Multiply 2.7 by 1000.

2.7 * 1000 = 2700
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3 years ago
HELPPPPPPPPPPPPPPPP!!!!!!
timofeeve [1]

Answer:

15

Step-by-step explanation:

It is a right angle so it will add up to 90 degrees.

This is the equation:

x=90-19-56

x=15

4 0
3 years ago
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I NEED HELP PLEASE<br> Find the measure of ZD.
IgorLugansk [536]

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2 years ago
The transformation T = [0 1 -1 0] is applied the figure below.
Margaret [11]

Answer:

<h2>a.) reflect across x-axis</h2>

Step-by-step explanation:

The transformation described is about multiplying the vertical value by -1:

(x,y) \implies (x,-y)

That means all vertical coordinates will change to the opposite side, but all horizontal coordinates will maintain at the same coordinate.

As a result, we'll have a reflection across the x-axis, because the y coordinates were transformed.

Therefore, the right answer is A.

4 0
3 years ago
A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function
dexar [7]
Here's our equation.

h=-16t+64t+3

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


8 0
4 years ago
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