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3 years ago

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Evaluate 2(x+ 1 - 3 when x = 6 O Ο Α. 11 OB. 5 O C. 8 ( D. 10

1 answer:

Mashutka [201]3 years ago

5 0

$\implies {\blue {\boxed {\boxed {\purple {\sf {A. \:11}}}}}}$

$\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}$

$2 \: ( \: x + 1 \: ) - 3$

Plugging in the value " $x\:=\:6$ " in the above expression, we have

$= 2 \: ( \: 6 + 1 \: ) - 3$

$= 2 \: ( \: 7 \: ) - 3$

$= 14 - 3$

$= 11$

<h3><u>Note</u>:-</h3>

$\sf\red{PEMDAS\: rule.}$

P = Parentheses

E = Exponents

M = Multiplication

D = Division

A = Addition

S = Subtraction

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}$

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Which of the following equations has the same solutions as the equations 3x+2y=-12

Marianna [84]

B, C, E, F

<h2>Explanation:</h2>

In this problem, we have the following equation of a line written in Standard form:

$3x+2y=-12$

So we need to choose the equations that has the same solutions as this equation. If we multiply each equation by a constant term and find the same solution as the given equation, then they will have the same solution. This is only true fro options B, C, E and F.

Option B.

$-3x-2y= 12 \\ \\ Multiply \ both \ sides \ by \ -1: \\ \\ (-1)(-3x-2y)=(-1)(12) \\ \\ 3x+2y=-12$

So we get the same given equation. This option is valid!

Option D.

$15x+10y= -60 \\ \\ Multiply \ both \ sides \ by \ 1/5: \\ \\ (\frac{1}{5})(15x+10y)=(\frac{1}{5})(-60) \\ \\ 3x+2y=-12$

So we get the same given equation. This option is valid!

Option E.

$6x+2y= -12 \\ \\ Multiply \ both \ sides \ by \ 1/2: \\ \\ (\frac{1}{2})(6x+2y)=(\frac{1}{2})(-24) \\ \\ 3x+2y=-12$

So we get the same given equation. This option is valid!

Option F.

$1.5x+y= -6 \\ \\ Multiply \ both \ sides \ by \ 2: \\ \\ (2)(1.5x+y)=(2)(-6) \\ \\ 3x+2y=-12$

So we get the same given equation. This option is valid!

For the other options, there is no any constant term that multiplies both sides of the equation and gives us the same given equation.

<h2>Learn more:</h2>

About this subject: brainly.com/question/8810210

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3 years ago

The number of new cars purchased in a city can be modeled by the equation C=20t to the 2nd power, where C is the number of new c

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Answer:

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Step-by-step explanation:

Set up the equation:

Since C(t) gives the number of cars purchased in the t-th year after 1998, then make the number of cars equal to 15 000 and solve for t - the year:

20t^2 = 15000

t^2 = 750

t = $\sqrt{750} = \sqrt{10} \cdot \sqrt{25} \cdot \sqrt{3} = 5\sqrt{30}$

The year will be simply 1998 + 5 \sqrt{30}

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