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Westkost [7]
3 years ago
11

1. A boy needs to be at school by 8:15 am. If it takes him 15

Mathematics
2 answers:
GenaCL600 [577]3 years ago
5 0

Answer:

7:00 am

Step-by-step explanation:

7:00 am

pshichka [43]3 years ago
4 0

Answer:

He would need to get up at 7:00am

Step-by-step explanation:

He would need to get up at 7:00 am because if you add all the times together which would be 15+20+25+15=75 so he would take 1 hour and 15 mins to get ready before school therefor he would be at school exactly at 8:15 am.hope it helps you!

<h2>        </h2>
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ella [17]

Answer:

I think the answer is m= -91

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If a tower casts a shadow 11 feet long the distance from the top of the tower to the end of the shadow is 61 feet .. How many fe
Svetach [21]
Use\ the\ Pythagoras\ Theorem:\\\\H^2+11^2=61^2\\\\H^2+121=3721\ \ \ \ |subtract\ 121\ from\ both\ sides\\\\H^2=3600\\\\H=\sqrt{3600}\\\\\boxed{H=60}\\\\Answer:Tower\ is\ 60\ feet\ tall.Answer here

8 0
3 years ago
A 15m long ladder reached a window 9m high from the ground on placing it against a wall at a distance x. Find the distance of th
Ilya [14]

Answer:

12 meters

Step-by-step explanation:

Given That :

Length of ladder = 15m

Height of wall =. 9 m

Using Pythagoras rule :

Adjacent² = hypotenus² - opposite²

Adjacent² = 15² - 9²

Adjacent² = 225 - 81

Adjacent² = 144

Adjacent = sqrt(144)

Adjacent, x = 12

8 0
3 years ago
A parabola can be represented by the equation x2 = -20y.
aliya0001 [1]

Answer:

\mathrm{Parabola\:focus\:given}\:x^2=-20y:\quad \left(0,\:-5\right)

Step-by-step explanation:

Given the equation

x2 = -20y

A parabola is the locus of points such that the distance to a point the focus equals the distance to a line the directrix.

4p\left(y-k\right)=\left(x-h\right)^2 is the standard equation for an up-down facing parabola with vertex at (h, k), and a focal length |p|.

so

x^2=-20y

\mathrm{Switch\:sides}

-20y=x^2

\mathrm{Factor\:}4

4\cdot \frac{-20}{4}y=x^2

4\left(-5\right)y=x^2

\mathrm{Rewrite\:as}

4\left(-5\right)\left(y-0\right)=\left(x-0\right)^2

\left(h,\:k\right)=\left(0,\:0\right),\:p=-5

Parabola is symmetric around the y-axis and so the focus lies a distance\ p from the center (0, 0) along the y-axis.

\left(0,\:0+p\right)

=\left(0,\:0+\left(-5\right)\right)

\mathrm{Refine}

\left(0,\:-5\right)

Therefore,

\mathrm{Parabola\:focus\:given}\:x^2=-20y:\quad \left(0,\:-5\right)

Please check the attached figure too.

6 0
3 years ago
Read 2 more answers
Use the tangent ratio to find x in the picture attached.
Sedbober [7]
Sry that I can’t answer but I hope u have a great day
Yxucivb:)
4 0
3 years ago
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