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Alexeev081 [22]
3 years ago
7

Every day, Susan exercises in the morning. She travels a total of 4 miles. She starts off with a half-hour walk at a pace of 2 m

iles per hour. She then finishes while jogging at a pace of 6 miles per hour. Write an equation that represents this situation, and use it to find the number of hours Susan jogs at the increased speed.
Mathematics
1 answer:
Anna [14]3 years ago
7 0

Let's assume

the number of hours Susan jogs at the increased speed is 't' hours

Since, increased speed is 6 miles per hour

after 't' hours , she travles with increased speed is 6*t miles

She starts off with a half-hour walk at a pace of 2 miles per hour

so, she travels distance

=0.5*2 miles

=1 miles

She travels a total of 4 miles

total distance is

=6t+1

now, we can set them equal

6t+1=4

now, we can solve for t

Subtract both sides by 1

6t+1-1=4-1

6t=3

Divide both sides by 6

we get

t=0.5

so, the number of hours Susan jogs at the increased speed is 0.5 hrs.........Answer


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Find the 7th term in the sequence -10,-6,-2,2...
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1st term = -10

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The reason how I got 14 for the 7th term is because, i added 4 to each term.


Hope this helps!

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SOLUTION

Consider the property, opposite angle of a cyclic quadrilateral are supplementary

hence

\angle NMP+\angle NOP=180^0

Recall that

\begin{gathered} \angle NMP=\angle M=(8x-24)^0 \\ \text{and } \\ \angle NOP=\angle O=(4x)^0 \end{gathered}

Hnece, we have that

\begin{gathered} \angle M+\angle O=180^0(opposite\text{ angles of a cyclic quadrilateral)} \\ (8x-24)^0+4x^0=180^0 \end{gathered}

Then

\begin{gathered} 8x-24+4x=180 \\ 8x+4x-24=180 \\ 12x-24=180 \\ \text{Add 24 t o both sides } \\ 12x-24+24=180+24 \\ 12x=204 \\ \text{divide both sides by 12} \\ \frac{12x}{12}=\frac{204}{12} \\ \text{Then} \\ x=17 \end{gathered}

Hence

x=17

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\begin{gathered} \angle NOP=(4x)^0 \\ \text{substitute the value of x, we have } \\ \angle NOP=4\times17=68^0 \\ \text{Hence } \\ \angle NOP=68^0 \end{gathered}

Therefore

The measure of angle NOP is 68⁰

Answer; 68⁰ (The fourth Option )

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