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kodGreya [7K]
3 years ago
13

Which expression represents the volume of the pyramid? One-fifth(5.2)h cm3 StartFraction 1 Over 5 h EndFraction(5.2)h cm3 One-th

ird(5.2)h cm3 StartFraction 1 Over 3 h EndFraction(5.2)h cm3
Mathematics
2 answers:
elixir [45]3 years ago
8 0

Answer:

its c

Step-by-step explanation:

anyanavicka [17]3 years ago
7 0

Answer:

1/3(5.2)h cm³

Step-by-step explanation:

A solid right pyramid has a regular hexagonal base with an area of 5.2 cm2 and a height of h cm. Which expression represents the volume of the pyramid?

One-fifth(5.2)h cm3 StartFraction 1 Over 5 h EndFraction(5.2)h cm3

One-third(5.2)h cm3 StartFraction 1 Over 3 h EndFraction(5.2)h cm3

Volume of the pyramid = 1/3 × area × height

Area = 5.2 cm²

Height = h cm

Volume of the pyramid = 1/3 × 5.2 cm² × h cm

= 1/3(5.2)h cm³

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A ball is thrown into the air with a speed of 32 feet per second. The function h=32t - 16t^2 models the height of the ball after
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Answer:

t = 1 s

Step-by-step explanation:

The function models the height of the ball after t seconds is given by :

h=32t - 16t^2

We need to find how many seconds does it take for the ball to reach its maximum height.

For maximum height put dh/dt = 0

\dfrac{dh}{dt}=\dfrac{d}{dt}(32t - 16t^2)\\=32-32t

So,

32-32t = 0

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3 years ago
Can someone help me with this Geometry question, thanks!
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3 years ago
Read 2 more answers
Let theta be an angle in quadrant II such that cos theta = -2/3
Iteru [2.4K]

Answer:

So we have \csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}.

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given \cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}}).

So to find the opposite we will just use the Pythagorean Theorem.

a^2+b^2=c^2

(2)^2+b^2=(3)^2

4+b^2=9

b^2=5

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Now to find \csc(\theta) and \tan(\theta).

\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}.

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So \csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}.

And now \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}.

So we have \csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}.

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3 years ago
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The sum of five h and twice g is equal to 23
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