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3 years ago

What is 2(5+8)+101+101

Mathematics

2 answers:

Andre45 [30]3 years ago

3 0

2(5+8)+101+101
Step 1: 2(13)+(101+101) =
Step 2:26+202=
Final Answer: 228

umka2103 [35]3 years ago

3 0

The answer to your question is 228...

2(5+8)+101+101 - add (5+8)
2(13)+101+101 - multiply 2(13)
26+101+101 - Add 26,101,and 101
228

Hope this helps! :)

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mina [271]

Answer:

y inversely proportional to 1/x

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4=k/16

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y=8

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3 years ago

Can someone give me the answers plz:)

Alborosie

Remark

There is no short way to do this problem and no obvious way to get the answer other that to solve each part.

Solve

$\dfrac{x + 1.6}{2} =\text{x + 0.1}$ Multiply by 2

x + 1.6 = 2(x + 0.1) Remove the brackets

x + 1.6 = 2x + 0.1*2

x + 1.6 = 2x + 0.2 Subtract x from both sides

1.6 = x + 0.2 Subtract 0.2 from both sides

1.6 - 0.2 = x

1.4 = x

Circle A

Subtract 2x from both sides.

3x - 2x = 1.4

Circle B

Remove the brackets.

4x + 6 = 2x - 6 Add 6 to both sides

4x + 12 = 2x Subtract 4x from both sides.

12 = -2x Divide by - 2

12/-2 = x

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I'm going to be very scant in my solution of this. You can fill in the steps.

3x = 4.2

x = 4.2/3

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Circle D

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Is 3.50s+5.00L less than or equal to 80 a system of linear inequalities

kirill [66]

Answer:

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Answer:

complementary angles

Step-by-step explanation:

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How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

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