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Vladimir79 [104]
2 years ago
7

What is the area of this quadrilateral?

Mathematics
1 answer:
vagabundo [1.1K]2 years ago
5 0

Answer:

D

Step-by-step explanation:

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An object is launched with a velocity at 64 feet per second from a platform 30 feet high. the function H(t)= -16t2+64t+30 repres
PolarNik [594]
To find the maximum or minimum value of a function, we can find the derivative of the function, set it equal to 0, and solve for the critical points.

H'(t) = -32t + 64

Now find the critical numbers:

-32t + 64 = 0

-32t = -64

t = 2 seconds

Since H(t) has a negative leading coefficient, we know that it opens downward. This means that the critical point is a maximum value rather than a minimum. If we weren't sure, we could check by plugging in a value for t slightly less and slighter greater than t=2 into H'(t):

H'(1) = 32

H'(3) = -32

As you can see, the rate of change of the object's height goes from increasing to decreasing, meaning the critical point at t=2 is a maximum.

To find the height, plug t=2 into H(t):

H(2) = -16(2)^2 +64(2) + 30 = 94

The answer is 94 ft at 2 sec.
4 0
3 years ago
Shell charges $2.23 dollars per gallon of gas for refueling vehicles. Assume the number of gallons is your independent variable
sattari [20]
Y=$2.23x
Is there more to the question?
5 0
3 years ago
Three of these expressions give the distance between points A and B on the number line. Which expression does NOT?
NARA [144]

answer: was willing to help but you should’ve added a picture or added the choices there was to pick from because there isn’t much info.

7 0
3 years ago
What is the length of BC , rounded to the nearest tenth?
Arte-miy333 [17]

Step 1

In the right triangle ADB

<u>Find the length of the segment AB</u>

Applying the Pythagorean Theorem

AB^{2} =AD^{2}+BD^{2}

we have

AD=5\ units\\BD=12\ units

substitute the values

AB^{2}=5^{2}+12^{2}

AB^{2}=169

AB=13\ units

Step 2

In the right triangle ADB

<u>Find the cosine of the angle BAD</u>

we know that

cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}

Step 3

In the right triangle ABC

<u>Find the length of the segment AC</u>

we know that

cos(BAC)=cos (BAD)=\frac{5}{13}

cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}

\frac{5}{13}=\frac{AB}{AC}

\frac{5}{13}=\frac{13}{AC}

solve for AC

AC=(13*13)/5=33.8\ units

Step 4

<u>Find the length of the segment DC</u>

we know that

DC=AC-AD

we have

AC=33.8\ units

AD=5\ units

substitute the values

DC=33.8\ units-5\ units

DC=28.8\ units

Step 5

<u>Find the length of the segment BC</u>

In the right triangle BDC

Applying the Pythagorean Theorem

BC^{2} =BD^{2}+DC^{2}

we have

BD=12\ units\\DC=28.8\ units

substitute the values

BC^{2}=12^{2}+28.8^{2}

BC^{2}=973.44

BC=31.2\ units

therefore

<u>the answer is</u>

BC=31.2\ units

8 0
3 years ago
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I think it’s 6721 marbles in the larger jar.
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