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3 years ago

50 percent of 620 written as a percent

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

5 0

50% of 620 is half of 620.

Writing 50% of 620 is 50%. (Or 310 if you wanted to know half of 620)

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Read 2 more answers

What is the ratio of the leg opposite the angle to the leg adjacent to the<br> angle.

vichka [17]

Answer:

tangent (angle) or tan (angle)

Step-by-step explanation:

tangent (angle) = the leg opposite the angle / the leg adjacent to the angle

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2 years ago

A student who is training for a distance race jogged for thirty minutes and covered a distance of 7.5 km. What is the average sp

Tju [1.3M]

Answer:

<h2>Average speed is 15km/h</h2>

Step-by-step explanation:

Step one:

given data:

total distance =7.5km

time taken= 30 min= 0.5 hours

we know that the average speed is the average distance divided by time taken

Step two:

mathematically expressed as

speed=distance/time

substituting we have

speed=7.5/0.5

speed=15km/hour

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3 years ago

Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-

djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

$A(x_{1}, y_{1}})=(-20, 25)$

$B(x_{2}, y_{2}})=(30, 25)$

$C(x_{3}, y_{3}})=(30, -10)$

$D(x_{4}, y_{4}})=(-20, -10)$

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

$d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

$AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}$

$AB=\sqrt{(30+20)^{2}}$

$AB=\sqrt{(50)^{2}$

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

$d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$

$BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}$

$BC=\sqrt{(-35)^{2}}$

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

$d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}$

$DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}$

$DC=\sqrt{(30+20)^{2}}$

$DC=\sqrt{(50)^{2}}$

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

$d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}$

$AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}$

$AD=\sqrt{(-20+20)^{2}+(-35)^{2}}$

$AD=\sqrt{(-35)^{2}}$

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

7 0

3 years ago