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2 years ago

Solve this inequality for x: -46 - 8x > 22 A. X -8.5 D. X > 8.5

Mathematics

1 answer:

Juli2301 [7.4K]2 years ago

6 0

The answer would be “X < -8.5”
Have a good day:)

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Please helpp (exponents)

pshichka [43]

Answer:

the answer is 27

Step-by-step explanation:

for the show your work all you have to do is write 3x3x3=27

5 0

2 years ago

Molly was on a long 136 mile road trip. The first part of the trip there was lots of traffic, she only averaged 16 mph. The seco

Mazyrski [523]

Answer:

In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

Explanation:

Given that the road trip was 136 miles;

$d=136$

The first part of the trip there was lots of traffic, she only averaged 16 mph;

$v_1=16$

The second part of the trip there was no traffic so she could drive 44 mph;

$v_2=44$

She traveled for a total of 5 hours;

$t=5$

let x represent the time in traffic when she traveled at 16 mph

$t_1=x$

the time the traffic is clear would be;

$t_2=t-t_1=5-x$

Recall that distance equals speed multiply by time;

$d=v_1t_1_{}_{}^{}+v_2t_2$

substituting the values;

$136=16x+44(5-x)$

solving for x;

$\begin{gathered} 136=16x+220-44x \\ 44x-16x=220-136 \\ 28x=84 \\ x=\frac{84}{28} \\ x=3 \end{gathered}$

So;

$\begin{gathered} t_1=3\text{ hours} \\ t_2=5-x=5-3=2 \\ t_2=2\text{ hours} \end{gathered}$

Therefore, In traffic, she drove for 3 hours

and After the traffic cleared she drove for 2 hours.

7 0

11 months ago

What is the length of the line segment with endpoints (−3,8) and (7, 8) ? Enter your answer in the box

ki77a [65]

Length = √(y2-y1)² + (x2-x1)²
l = √(8-8)² + (7+3)²
l = √10²
l = 10

In short, Your Answer would be 10 units

Hope this helps!

6 0

3 years ago

Read 2 more answers

If x / 3 equals 6 then x equals

sleet_krkn [62]

Answer: X = 18

Step-by-step explanation:

Crossmultiply

8 0

2 years ago

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

7 0

2 years ago