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anygoal [31]
3 years ago
13

How would I write the equation for this table?

Mathematics
1 answer:
Rudiy273 years ago
6 0

Answer:

Step-by-step explanation:

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Suzanna walked two over seven of a mile in two over five of an hour. What is her unit rate in miles per hour? two sevenths over
Lorico [155]

Answer:

5/7 mile per hour.

Step-by-step explanation:

Divide 2/7 by 2/5 to get to 0.714. After that, convert the decimal to a fraction and you get your answer.



Hope this helped!!! <3 :)

5 0
3 years ago
Does it matter which way you subtract in distance
lubasha [3.4K]
Yes it does matter which way you subtract


5 0
3 years ago
Eric drove 248 miles in 4 hours. If he continues to travel at the speed, how far will he go in 7 hours and 30 minutes?
qaws [65]
248/4=62
62•7 1/2=465 miles

He will go 465 miles in 7 hours and 30 minutes
6 0
3 years ago
Read 2 more answers
HELPPPP PLEASEEE IM BEGGING SOMEONE PLEASE PLEASE HELP
pychu [463]

Given:

The figure of a rhombus QRST.

To find:

A. The value of x.

B. The measure of angle RQP.

Solution:

A. We need to find the value of x.

We know that the diagonals of a rhombus are perpendicular bisectors. It means the angles on the intersection of diagonals are right angles.

m\angle RPS=90^\circ                       [Right angle]

(5x+15)^\circ=90^\circ

(5x+15)=90

5x=90-15

5x=75

Divide both sides by 5.

x=\dfrac{75}{5}

x=15

Therefore, the value of x is 15.

B. We need to find the measure of angle RQP.

From the given figure, it is clear that

m\angle RQP=(2x+3)^\circ

Putting x=15, we get

m\angle RQP=(2(15)+3)^\circ

m\angle RQP=(30+3)^\circ

m\angle RQP=33^\circ

Therefore, the measure of angle RQP is 33 degrees.

8 0
2 years ago
Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i
dsp73

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s

Also,

v=r\omega

Where

\omega is the angular speed

So,

\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

8 0
3 years ago
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