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3 years ago

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What is 8 to the second power open parentheses 3x - 2 closed parentheses + 5 = 56. solve for x

1 answer:

Ratling [72]3 years ago

8 0

8² (3x -2) + 5 = 56 I do the exponents first
64 (3x -2) + 5 = 56 I distribute the 64 over the 3x-2
192x - 128 + 5 = 56 I combined liked terms (-128+5)
192x -123 = 56 I add 123 to both side
192x - 123 +123= 56 + 123 Finally I divide both sides by 192
<u>192x</u> = <u>179
</u>192 192

x = $\frac{179}{192}$

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Graph -7x+5y=35. khan acadmy forms of linear equations question. pls show work (10 points)

a_sh-v [17]

Answer:

Below

Step-by-step explanation:

● -7x + 5y = 35

Add 7x to both sides

● -7x +7x + 5y = 35+7x

● 5y = 7x + 35

Divide both sides by 5

● 5y/5 = (7x+35)/5

● y = 1.4x + 7

The graph of the function:

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3 years ago

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What is the equation of the line that passes through the point (-5,5)and has a slope of -1

erastova [34]

Answer:

<h2> y = - x </h2>

Step-by-step explanation:

The point-slope form of the equation is y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope:

m = -1

(-5, 5) ⇒ x₀ = -5, y₀ = 5

The point-slope form of the equation:

y - 5 = -1(x + 5)

So:

y - 5 = -x - 5 {add 5 to both sides}

y = -x ← the slope-intercept form of the equation

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3 years ago

Which ratio shows tangent if angle A.

Alexandra [31]

Answer:

8/15

Step-by-step explanation:

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3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

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Masteriza [31]

The money altogether is 8.00$

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