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2 years ago

What is the answer to 4x + 27

Mathematics

1 answer:

Alla [95]2 years ago

7 0

Answer:

Step-by-step explanation:

4x+27

---- ----

4 4

x=6.75 or x=7

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If the following cylinders shown below are similar, what is the value of x? A.3 B.5<br> C.10 D. 20

Inessa05 [86]

Answer:

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2 years ago

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To which linear equations is the coordinate a solution?

never [62]

The linear equation y = -x - 2 and y = 2x + 13 form the solution (-5, 3)

<h3>Linear equation</h3>

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the rate of change and b is the y intercept.

Given the coordinate (-5, 3) as a solution.

The linear equation y = -x - 2 satisfies this this equation because 3 = -(-5) -2

Also, the linear equation y = 2x + 13 satisfies this this equation because 3 = 2(-5) + 14

Find out more on Linear equation at: brainly.com/question/14323743

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1 year ago

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Factor the expression using gcf of 45+30

Korolek [52]

Answer:

15(3+2)

Step-by-step explanation:

45/15=3

30/15=2

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2 years ago

a rental car cost $50 for one day plus an additional 0.25 per mile. what is the cost of renting a car for one day, and driving i

Natali5045456 [20]

Answer:

you answer is 68.75

Step-by-step explanation:

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2 years ago

Divide.<br><br> 6√3 cis(7π/6) ÷ 3√5 cis(π/3)

Nesterboy [21]

$\dfrac{6\sqrt 3~ \text{cis} \left( \dfrac{7 \pi} 6 \right)}{3 \sqrt 5 ~\text{cis} \left( \dfrac{\pi}{3}\right)}\\\\\\=\dfrac{2\sqrt 3 \cdot e^{i\tfrac{7\pi}{6}}}{\sqrt 5\cdot e^{i \tfrac{\pi}{3}}}\\\\\\=\dfrac{2 \sqrt 3}{\sqrt 5} \cdot \left(e^i \right)^{\tfrac{7\pi}{6} - \tfrac{\pi }{3}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~;\left[\text{cis}~ \theta = e^{i \theta}\right]\\\\\\=\dfrac{2 \sqrt{3}}{\sqrt 5} \cdot \left(e^i\right)^{\tfrac{5\pi}6}\\\\\\=\dfrac{2\sqrt{3}}{\sqrt 5}\cdot e^{i\tfrac{5 \pi}6}\\\\\\$

$=\dfrac{2\sqrt{3}}{\sqrt 5} \left[ \cos \left(\dfrac{5\pi}{6}\right) + i \sin \left(\dfrac{5\pi}{6}\right) \right]~~~~~~~~~~~~~~~~~~;\left[e^{i\theta} = \cos \theta + i \sin \theta} \right]\\\\\\=\dfrac{2\sqrt{3}}{\sqrt 5}\left[ \cos \left(2 \times \dfrac{\pi}2 -\dfrac{\pi}{6} \right) + i\sin \left(2 \times \dfrac{\pi}2 -\dfrac{\pi}{6} \right) \right]\\\\\\=\dfrac{2\sqrt{3}}{\sqrt 5} \left[ -\cos \left(\dfrac{\pi}{6}\right) + i \sin \left(\dfrac{\pi}{6}\right) \right]\\\\\\$

$=\dfrac{2\sqrt{3}}{\sqrt 5}\left( -\dfrac {\sqrt{3}}{2} + i \cdot \dfrac 12 \right)\\\\\\=-\dfrac{3}{\sqrt 5} + i \dfrac{\sqrt 3}{\sqrt 5}$

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1 year ago