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3 years ago

Solve for y. 4/9(y-3)=g

1 answer:

6 0

4/9(y-3) = g....multiply both sides by 9/4
9/4(4/9)y-3 = 9/4g
y - 3 = 9/4g....now add 3 to both sides
y - 3 + 3 = 9/4g + 3
y = 9/4g + 3

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This is really easy! pic below..

KATRIN_1 [288]

The answer is A.
I think so because each one of them agrees with 20x and 15

5 0

3 years ago

Find the equation of the line tangent to the graph of f(x)=(lnx)^(4)at x=10

Diano4ka-milaya [45]

Hello,

Step-by-step explanation:

$f(x) = ln(x) {}^{4}$

$(ln(u)') = \frac{u'}{u}$

$f'(x) = \frac{4ln {}^{} (x) {}^{3} }{x}$

$f'(10) = \frac{4ln {}^{} (10) {}^{3} }{10} = \frac{12ln(x)}{x}$

$f(10) = ln(10) {}^{4}$

$y = \frac{12ln(x)}{x} (x - 10) + 4ln(10)$

$y = f'(a)(x - a) + f(a)$

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2 years ago

Sixty-three is 70% of what number?

Margaret [11]

70 =63
100 ?
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3 0

3 years ago

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

5 0

3 years ago

Could someone pls answer

cricket20 [7]

The answer should be 10

4 0

1 year ago