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Alex_Xolod [135]
3 years ago
13

A student fundraiser earned 30% profit on t-shirts sold. If the students sold $500 in t-shirts, how much of the money was profit

Mathematics
2 answers:
Lady_Fox [76]3 years ago
6 0
All you have to do is .30 x 500 and get 150
Triss [41]3 years ago
4 0
I am assuming that this question is asking much 30% of 500 is, which is 150.
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if H(x) is x²

Step-by-step explanation:

3x + 2 + 2x - 1 +  {x}^{2}  =  \\   3x + 2x = 5x \\  {x}^{2}  + 5x - 1

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Sarah saving account balance change by $35 one week and by negative $67 to next week. Which amount represents the greatest chang
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I think it is negative $65.
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Will mark Brainliest if someone can help me!!!!! (Idk if Prinicipal is right 2-5)
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\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1500\\ r=rate\to r\%\to \frac{r}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1,2,3,4,5 \end{cases}

\bf A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 1}\implies A=1500(1.04)^1\implies A\approx 1560\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 2}\implies A=1500(1.04)^2\implies A\approx 1622.4\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 3}\implies A=1500(1.04)^3\implies A\approx 1687.3\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 4}\implies A=1500(1.04)^4\implies A\approx 1754.79\\\\\\A=1500\left(1+\frac{0.04}{1}\right)^{1\cdot 5}\implies A=1500(1.04)^5\implies A\approx 1824.98

6 0
3 years ago
A ribbon measure 4.5 meters express length to millimeters<br>​
GREYUIT [131]

Answer:

4500 millimeters

Step-by-step explanation:

6 0
3 years ago
LOTS OF POINTS GIVING BRAINLIEST I NEED HELP PLEASEE
Sidana [21]

Answer:

Segment EF: y = -x + 8

Segment BC: y = -x + 2

Step-by-step explanation:

Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.

Upon observing the given diagram, we can infer the following corresponding sides:

\displaystyle\mathsf{\overline{BC}\:\: and\:\:\overline{EF}}

\displaystyle\mathsf{\overline{BA}\:\: and\:\:\overline{ED}}

\displaystyle\mathsf{\overline{AC}\:\: and\:\:\overline{DF}}

We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.

<h2>Slope of Segment BC:</h2>

In order to solve for the slope of segment BC, we can use the following slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}  }

Use the following coordinates from the given diagram:

Point B:  (x₁, y₁) =  (-2, 4)

Point C:  (x₂, y₂) = ( 1,  1 )

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}

<h2>Slope of Segment EF:</h2>

Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:

Point E:  (x₁, y₁) =  (4, 4)

Point F:  (x₂, y₂) = (6, 2)

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}

Our calculations show that segment BC and EF have the same slope of -1.  In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.  

Since segments BC and EF have the same slope, then it means that  \displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}.

<h2>Slope-intercept form:</h2><h3><u>Segment BC:</u></h3>

The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.

Using the slope of segment BC, m = -1, and the coordinates of point C, (1,  1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>

y = mx + b

1 = -1( 1 ) + b

1 = -1 + b

Add 1 to both sides to isolate b:

1 + 1 = -1 + 1 + b

2 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.

Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:

⇒  y = -x + 2.

<h3><u /></h3><h3><u>Segment EF:</u></h3>

Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>

y = mx + b

4 = -1( 4 ) + b

4 = -4 + b

Add 4 to both sides to isolate b:

4 + 4 = -4 + 4 + b

8 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.

Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:

⇒  y = -x + 8.

8 0
2 years ago
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