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pickupchik [31]
3 years ago
12

Which graph représents an exponential function?

Mathematics
1 answer:
Rama09 [41]3 years ago
5 0

Answer:

Photographs and graph is uncomfortable to understand what you have right can you please edit the type of photograph photo or anything please please and I keep some more points in 5 points no one is going to us answer your question

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Write <img src="https://tex.z-dn.net/?f=cot%5E%7B2%7D%20x%20%281%2Btan%5E%7B2%7D%20x%29" id="TexFormula1" title="cot^{2} x (1+ta
ASHA 777 [7]

Step-by-step explanation:

From Trig

1 + tan²x = Sec²x

Also

Sec²x = 1/cos²x

Now

Cot²x = 1/tan²x = cos²x/sin²x

Putting all together

Cot²x(1+tan²x)

= Cot²x(sec²x)

= cos²x/sin²x(1/cos²x)

cos²x on the numerator and that one the denominator cancels out

we have

= 1/sin²x

From Trig

1/sin²x = cosec²x

So Our Answer = cosec²x.

7 0
3 years ago
what is the exact circumference of the base of a cone that has a volume of 1014 pi in^3 and a height of 18 in.
Evgen [1.6K]
\bf \textit{volume of a cone}\\\\&#10;V=\cfrac{\pi r^2 h}{3}\quad &#10;\begin{cases}&#10;r=radius\\&#10;h=height\\&#10;-----\\&#10;V=1014\pi \\&#10;h=18&#10;\end{cases}\implies 1041\pi =\cfrac{\pi \cdot r^2\cdot 18}{3}&#10;\\\\\\&#10;1041\pi =6\pi r^2\implies \cfrac{1041\pi }{6\pi }=r^2\implies \cfrac{347}{2}=r^2\implies \boxed{\sqrt{\cfrac{347}{2}}=r}\\\\&#10;-------------------------------\\\\&#10;\textit{circumference of a }\stackrel{cone's}{circle}\\\\&#10;C=2\pi r\qquad \qquad C=2\pi \left( \boxed{\sqrt{\cfrac{347}{2}}} \right)
8 0
3 years ago
Which statement about this system of equations is true?
Lunna [17]
Parallel lines have no solution
8 0
3 years ago
a recipe calls for 2/3 of a cup of sugar per batch. elena used 5 and 1/3 cups of sugar to make multiple batches of cookies. How
Leto [7]

Answer:

8

Step-by-step explanation:

5 and 1/3 is also 16/3.    Then, 2x8 equals 16

5 0
3 years ago
Read 2 more answers
What is the inverse of this function? (Function in photo)
valkas [14]

Given:

The function is

f(x)=-\dfrac{1}{2}\sqrt{x+3},x\geq -3

To find:

The inverse of the given function.

Solution:

We have,

f(x)=-\dfrac{1}{2}\sqrt{x+3}

Put f(x)=y.

y=-\dfrac{1}{2}\sqrt{x+3}

Interchange x and y.

x=-\dfrac{1}{2}\sqrt{y+3}

Isolate the variable y.

-2x=\sqrt{y+3}

(-2x)^2=y+3

4x^2-3=y

y=4x^2-3

Putting y=f^{-1}(x), we get

f^{-1}(x)=4x^2-3

For x\geq -3,

x+3\geq 0

\sqrt{x+3}\geq 0

-\dfrac{1}{2}\sqrt{x+3}\leq 0

f(x)\leq 0

It means for function f(x)\leq 0 and for inverse function x\leq 0 because the range of the function is the domain of inverse function.

Therefore, f^{-1}(x)=4x^2-3 for x\leq 0.

3 0
3 years ago
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