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3 years ago

Solve the equation 14=7(2x-4) using two different methods. Show your work. Which method do you prefer? Explain. need help thanks

Mathematics

2 answers:

Jet001 [13]3 years ago

5 0

Answer:

x=3

Step-by-step explanation:

14=7(2x-4)

14/7=2x-4

2=2x-4

2x=2+4

2x=6

x=6/2

x=3

------------

14=7(2x-4)

14=14x-28

14x=14+28

14x=42

x=42/14

x=3

vodka [1.7K]3 years ago

3 0

$14=7(2x-4)\\2=2x-4\\2x=6\\x=3\\\\14=7(2x-4)\\14=14x-28\\14x=42\\x=3$

I prefer the first method, because I don't do unnecessary multiplication and therefore operate on smaller numbers.

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A rectangular prism with a square base has height 12 ft. The volume of the prism is 3468 cubic feet. Determine the side length o

Varvara68 [4.7K]

Answer:

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5 0

3 years ago

PLEASE HELP :( ASAP

kenny6666 [7]

Answer:

A. 272 cm^2

Step-by-step explanation:

The area of the rectangle is 17cm * 7cm = 119 cm^2

The area of the right triangle is (25-7) * 17cm * 1/2 = 153cm^2

<em>Answer: 119+153 = 272 cm^2</em>

3 0

3 years ago

Read 2 more answers

A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1+

egoroff_w [7]

Answer:

$a=\frac{3\times 7}{7}+\frac{12}{7}$ (First step for obtaining a common denominator for the two fraction)

$a=\frac{33}{7}\text{ m/s}^2$

Step-by-step explanation:

$\text{Given Expression: }a=a_1+\frac{F}{m}$

where,

a is acceleration of an object.(Need to calculate)

$a_1=3.00\text{ m/s}^2$

$F=12.0\text{ Kg.m/s}^2$

$m=7.00\text{ kg}$

$\text{Substitute }a_1, \text{ F, and m into expression and we get}$

$a=3+\frac{12}{7}$

Now we will simplify above expression for a

First we make common denominator.

Common denominator is 7. So, we make both denominator 7. We multiply by 7 at top and bottom with 3. We get

$a=\frac{3\times 7}{7}+\frac{12}{7}$ (First step for obtaining a common denominator for the two fraction)

Now we combine the numerator

$a=\frac{21+12}{7}$

$a=\frac{33}{7}\text{ m/s}^2$

3 0

4 years ago

Read 2 more answers

Suppose f (x) = x^2. Find the graph of f(x+2)

NISA [10]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\\\
f(x)={{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
~~~~~~if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
~~~~~~if\ {{ D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

with that template in mind,

$\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}$

C = 2 B = 1 C/B = 2/1 or +2, horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).

8 0

3 years ago

State a true conclusion.

FromTheMoon [43]

Answer:

RQ + QS = RS

Hope this helps!

3 0

3 years ago