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3 years ago

Please help!!! Solve for f. d (e – f) = g

Mathematics

1 answer:

Orlov [11]3 years ago

3 0

$d(e - f)= g\\\\de-df=g\\\\df=de-g\\\\f=\dfrac{ed-g}{d}$

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If the circle below is cut from the square of plywood below, how many square inches of plywood would be left over?

Katyanochek1 [597]

Answer:

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3 years ago

A rectangle measures 15 1/4 feet by 8 11/15 feet. If the length and width are extended by 1 2/3 Feet each find the area of the n

zubka84 [21]

Answer:

New area is 175 42/45 feet ²

Step-by-step explanation:

Given data

Length l= 15 1/4

To proper fraction = 61/4 feet

Width w= 8 11/15

To proper fraction = 131/15 feet

Extention = 1 2/3

To proper fraction = 5/3

Dimensions of new rectangle

Length =61/4+5/3 = 183+20/12

LCM = 12 = 203/12

Width = 131/15+5/3= 131+25/15

LCM = 15 = 156/15

Area = 203/12*156/15= =31668/180

=31668/180

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3 years ago

Which values are solutions to the inequality below? <br><br>Check all that apply.<br><br>x2 < 81

11111nata11111 [884]

A(6) and D(-4) are solutions to the equations

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5 0

3 years ago

Read 2 more answers

The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation

rodikova [14]

Answer: C) $y=sin^-1 x$

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point $(1,\frac{\pi}{2})$ ,

If this point satisfies the function,

For the function $y=cos^{-1} x$

If x = 1

$y=cos^{-1}1=0$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=cos^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=cot^{-1}x$

If x = 1

$y=cot^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cot^{-1}x$ ,

⇒ The graph of $y=cot^{-1}x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=sin^{-1} x$

If x = 1

$y=sin^{-1}1=\frac{\pi}{2}$

Thus, $(1,\frac{\pi}{2})$ is satisfying function $y=sin^{-1} x$ ,

⇒ The graph of $y=sin^{-1} x$ is passing through the point $(1,\frac{\pi}{2})$ .

For the function $y=tan^{-1}x$

If x = 1

$y=tan^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=tan^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$ .

Hence, Option C is correct.

3 0

3 years ago