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2 years ago

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To get to school from her house, Donita walks at a rate of 3 mph north along Troy Ave for 15 minutes. Then she walks at a rate o

f 3.75 mph east along Mesa St for 20 minutes. If she walks directly from her house to school, how fast must she walk to
arrive there in the same amount of time? Fill in the blanks, rounding answers to the nearest tenth.

1 answer:

Zielflug [23.3K]2 years ago

4 0

I mile = 60 / 3 = 20 minutes. Answer: The school is 1 mile from Emma's house. Lennox Obuong Algebra Tutor Nairobi, Kenya ...

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I’ll give a brainliest to whoever answers this ::::Find the equation of the line through the point A(4,6) and with a slope m=3

asambeis [7]

hi

line equation is : mx+p

so if a line goes throught A(4;6) and with m = 3 so you have 3 *(4) +P = 6

so : 12 +P = 6

P = 6-12

P = -6

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2 years ago

49/6 into a mixed number

mote1985 [20]

49/6 into a mixed number would be 8 1/6. This is because you would see how many times 6 can go into 49. Since 6 times 8= 48, you subtract that from 49. You are left with one. You put the number you multiplied 6 by as the whole number and the leftover one over six. I hope this helps!

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3 years ago

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Leno4ka [110]

The answer is 91 and the property used is the Identity Property of Addition.

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3 years ago

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Simplify the expression 3x(4x^4 - 5x)

Soloha48 [4]

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3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

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3 years ago