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Varvara68 [4.7K]
2 years ago
11

Marshall left home biking at a constant rate of 20 mph. One hour later, it looked like it may rain, so his friend, Brett, set ou

t in his car to pick him up. Brett drove at a constant rate of 60 mph. How many hours will Marshall have been riding his bike by the time Brett catches up to him?
A. 0.5
B. 1.0
C. 1.5
D. 2.0
Mathematics
1 answer:
Alexus [3.1K]2 years ago
6 0
First, we need to find how far ahead Marshall was.  Since he had been biking at 20 mph for one hour, he had gone 20 miles.

Next, we need to find how long it will take Brett to catch up to Marshall.  In order to do this, we need to find how much faster Brett is going than Marshall.  We do this by subtracting Marshall's speed from Brett's speed.

60 - 20 = 40.  So, Brett is catching up to Marshall at 40 mph.  Now, we figure out how long it will take for someone going 40 miles per hour to go 20 miles.  We find this by dividing 40 miles per hour by 20.  This is equal to 1/2 hour.  So, it will take Brett 0.5 hours to catch up to Marshall.  This is the same as A, so A is the correct answer.

We can check our answer by seeing how far Marshall and Brett will have gone.  Marshall will have been biking for 1.5 hours, so we multiply 20 * 1.5 = 30.  Marshall went 30 miles.

Brett drove for .5 hours at 60 mph, so he went 30 miles.  Since Brett and Marshall went the same distance, our answer is correct.
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Sofia has to order in rolls

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She has paid for 24 pieces (3 rolls) at $8 per roll

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In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
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Answer:

Perimeter  = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

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