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spayn [35]
3 years ago
13

Angela enjoys swimming and often swims at a steady pace to burn calories. At this pace, Angela can swim 1,700 meters in 40 minut

es. what is Angela's unit rate?
Mathematics
2 answers:
Alexxx [7]3 years ago
6 0
I think the answer is 

42.5 

not sure but hope this helps

yarga [219]3 years ago
4 0

Answer: Her unit rate is 42.5 m/min.

Step-by-step explanation:

Since we have given that

Distance she covered = 1700

Time taken = 40 minutes

So, we need to find the unit rate.

As we know that Speed=\dfrac{Distance}{Time}

Speed is given by

\dfrac{Distance}{Time}\\\\=\dfrac{1700}{40}\\\\=42.5\ m/min

Hence, her unit rate is 42.5 m/min.

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Aleks04 [339]

There is no answer? So how am I supposed to answer it?

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2 years ago
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Bradley bought 14 yellow highlighters. each highlighter cost $0.94.how much did Bradley spend?
Svetradugi [14.3K]
The answer would be 13.16$

0.94x14=13.16
6 0
3 years ago
Shira’s Shoes sold 875,000 pairs of sandals in June, which was 70% of the total number of shoes sold. How many shoes did the com
azamat

Answer:

1250000

Step-by-step explanation:

Let number of the total number of shoes sold be 'y'

y*(70/100)= 875000

70y = 875000*100

y = (875000*100)/70

y = 1250000

3 0
3 years ago
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3 moon + 1 star = 29 1 moon +1 moon = 15 Find value of 1 moon​
Montano1993 [528]

Hi!

I can help you with joy!

3 moons+1 star equals 29.

All right.

Actually, there are a few numbers that can be hidden behind the moon:

3*8=24 + 1 star (5) = 29

But 1 moon + 1 moon equals 15.

So one moon equals 7.5.

*GentleGirlie*

8 0
3 years ago
Find the inverse of the given​ matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3
BabaBlast [244]

Answer:

A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]

Step-by-step explanation:

We want to find the inverse of A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Subtract row 1 multiplied by 4 from row 2

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]

  • Add row 1 multiplied by 4 to row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]

  • Subtract row 2 multiplied by 2 from row 3

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]

  • Divide row 3 by −27

\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Add row 3 multiplied by 2 to row 1

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

  • Subtract row 3 multiplied by 11 from row 2

\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0
3 years ago
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