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3 years ago

Can you help me solve another equation

Mathematics

1 answer:

Katarina [22]3 years ago

5 0

わからない、ごめんなさい。ポイントが欲しかったです笑

Hope this helped :))

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Sylvia earns $7 per hour at her after school job. After working one week, she received a paycheck for $91.

makvit [3.9K]

Answer:

a.) $hours=13$

b.) $h\leq 15$

c.) $ $$105$

Step-by-step explanation:

a.) If she earns $7 'per' hour, this multiplication. After a week, we need to see how many hours she worked to earn $91, so this goes behind the equal sign. Let h be the unknown number of hours:

$7h=91$

Solve for h. Divide both sides by 7:

$\frac{7h}{7}=\frac{91}{7}\\\\h=13$

Sylvia worked 13 hours at a rate of $7 an hour.

b.) She can work at most 15 hours a week. At 'most' can also mean less than or equal to. Let h be hours:

$h\leq 15$

Sylvia can work less than or equal to 15 hours per week.

c.) If Sylvia can work 15 hours, plug in this value for h into an equation when she earns $7 per hour. Let the product of the hours and money equal t for total:

$t=7(15)\\t=105$

The most money Sylvia can earn in a week is $105.

Finito.

6 0

3 years ago

What are the solutions of 2x^2+3x-7=x^2+5x+39

sveticcg [70]

Answer:

${x}^{2} - 2x - 46$

Step-by-step explanation:

${2x}^{2} + 3x - 7 = {x}^{2} + 5x + 39$

Collect like terms

${2x }^{2} - {x}^{2} + 3x - 5x - 7 - 39$

${ x}^{2} - 2x - 46$

7 0

4 years ago

What is the median score? A. 85 B.87 C.89 D.91

777dan777 [17]

B. 87 because the median is when you rank the all the numbers smallest to biggest, and the number in the middle is the median

5 0

3 years ago

Read 2 more answers

Monica retiled her square kitchen for $338. If the company she used charged $2 per square foot, how wide is her kitchen?

zloy xaker [14]

169 square feet wide.

3 0

4 years ago

Read 2 more answers

Given cos(4x)+2 Find: Period, Shift, & Range

Yuki888 [10]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}\qquad$

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

$\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{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's see

$\bf \begin{array}{llccll}
cos(&4x&+0)&+2\\
&\uparrow &\uparrow &\uparrow \\
&B&C&D
\end{array} 
\\\\\\
period\qquad \cfrac{2\pi }{B}\implies \cfrac{2\pi }{4}\implies \cfrac{\pi }{2}
\\\\\\
\textit{horizontal/phase shift}\qquad \cfrac{C}{B}\implies \cfrac{0}{4}\implies 0\impliedby none
\\\\\\
\textit{vertical shift}\qquad D=+2\impliedby \textit{2 units up}$

now the range, how far up and down it goes on the y-axis

well, for the graph of cos(x), the range is, goes up to 1, down to -1, is all,

the midline is at 0

now, with a vertical shift of 2 upwards, it moves the midline by 2 units, used to be at 0, now is at y = 2, but the amplitude never changed, is goes up and down still one unit,

but with the midline at 2, goes up to 3 and down to 1, so the range is 3 ⩽ y ⩽ 1

4 0

4 years ago

Can you help me solve another equation​

Can you help me solve another equation