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LenKa [72]
3 years ago
13

William cycles at a sped of 15 miles per hour. He cycles 12 miles from home to school. If he increases his cycling speed ny 5 mi

les per hour how much faster will he arrive his school
Mathematics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:he would arrive 0.2 miles faster.

Step-by-step explanation:

Distance = speed × time

Time = distance/speed

William cycles at a sped of 15 miles per hour. He cycles 12 miles from home to school. This means that the time it takes William to get to school from home would be

12/15 = 0.8 hours

If he increases his cycling speed by 5 miles per hour, his new speed becomes

15 + 5 = 20 miles per hour

Therefore, the new time it takes William to get to school from home would be

12/20 = 0.6 hours

The difference in both times is

0.8 - 0.6 = 0.2 hours

Therefore, he would arrive 0.2 miles faster.

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The rule is Paranthesis First

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|7-(24÷3)|-

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Step-by-step explanation:

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In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, fo
Scorpion4ik [409]

Answer:

x=\frac{8}{3}\ cm

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

A=\frac{1}{2}(AB)(BC)

we have

A=17\ cm^2

AB=(3x-2)\ cm

BC=(x+3)\ cm

substitute the values

17=\frac{1}{2}(3x-2)(x+3)

34=(3x-2)(x+3)

34=3x^2+9x-2x-6

3x^2+7x-6-34=0

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The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

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x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

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substitute in the formula

x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}

x=\frac{-7(+/-)\sqrt{529}} {6}

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x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}

x=\frac{-7(-)23} {6}=-5

therefore

The solution is

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