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nikdorinn [45]
4 years ago
14

In order to reflect an exponential function across the

Mathematics
1 answer:
Anna71 [15]4 years ago
6 0

Answer:

take the reciprocal of the initial value "a"

Step-by-step explanation:

From my past learning my teacher have taught me this.

If incorrect let me know

Hope this helps

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The lateral surface area of cone A is equal to the lateral surface area of cylinder B.
Natasha2012 [34]

Answer:

TRUE

Step-by-step explanation:

Lateral area of cone is given by: πrl

where r is the radius and l is the slant height

Here r=r and l=2h

Hence, lateral area of cone A= π×r×2h

                                            =  2πrh

Lateral area of cylinder is given by: 2πrh

where r is the radius and h is the height

Lateral area of cylinder B=2πrh

Clearly, both the lateral areas are equal

Hence, the statement that:The lateral surface area of cone A is equal to the lateral surface area of cylinder B. is:

True

4 0
3 years ago
Select the correct answer. Which equation, when solved, gives 8 for the value of x? OA. +3 = =+14 OB. 5-9=31-12 OC. 21-2=r-4 OD.
34kurt

Answer:

B

Step-by-step explanation:

i did it and it’s right

8 0
3 years ago
Does 32.60 round up or round down
adelina 88 [10]

That all depends what place you want it rounded to.

Rounding to the nearest tenth, it becomes  32.6 .
Rounding to the nearest whole number, it becomes,  33 .
Rounding to the nearest ten or higher order of magnitude, it becomes zero.

3 0
4 years ago
In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc
Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°

Use sine rule,

\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}

Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Thus, the angle B is function of A is, B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Now find \frac{dB}{dA}

Differentiate implicitly the function \sin{B}=\sqrt{\frac{3}{2}}\sin{A} with respect to A to get,

\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}

And,

f'(A)=\frac{dB}{dA}=\sqrt{3} from part b

Therefore, the linear approximation at A=\frac{\pi}{4} is,

f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}

d)

Use part (c), when A=46°, B is approximately,

B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°

8 0
3 years ago
Mrs. McClure's famous peanut butter cookies call for 1 cup of peanut butter for every 1 2 of a cup of oil. Today, she wants to m
bezimeni [28]

Answer:

Amount of butter need for 1 cup of oil = 2 cup of butter

Step-by-step explanation:

Given:

Amount of butter need for 1/2 cup of oil = 1 cup

Find:

Amount of butter need for 1 cup of oil

Computation:

Amount of butter need for 1/2 cup of oil = 1 cup

1 cup butter = 1/2 cup oil

So,

Amount of butter need for 1 cup of oil = 1 / [1/2]

Amount of butter need for 1 cup of oil = 1 / [0.5]

Amount of butter need for 1 cup of oil = 2 cup of butter

6 0
3 years ago
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