MATHEMATICS HIGH SCHOOL

A 12-foot flagpole casts a nine-foot shadow. What is the measure from the top of the shadow to the top of the flag pole?

Answers

Answer 1
Answer:

Here is the answer OwO:

A 12-foot flagpole casts a nine-foot shadow. The measure from the top of the shadow to the top of the flag pole is 15.

Here is the explanation/work OwO:

Switch sides.

c^2=12^2+9^2

12^2=144

c^2=144+9^2

9^2=81

c^2=144+81

Add the numbers: 144 + 81 = 225

c^2=225

For x^2 = f(a) the solution is √f(a)

c = 15

Therefore, your answer will be 15.

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Sorry if I am late. Hope this helps!! ^^
Answer 2
Answer:

Answer:

The answer is 15 ^-^

Step-by-step explanation:

Related Questions

Serious answers only :) 30 points!
Factorizar1. (a+ b)^3 − 272. 2x^2+7x+3
Which expression correctly represents “three less than the product of a number and two, increased by five”?
What is the perimeter of abc with vertices A(-2,9) B(7,-3) and C(-2,-3) in coordinate plane
If x = 3 units, y = 13 units, and h = 7 units, then what is the area of the trapezoid shown above?

The larger nail is a dilation of the smaller nail.Select all statements that are true.

Answers

Answer: the scale factor is 3/4

Step-by-step explanation:

Answer:

whats the rest of the answer?

Step-by-step explanation:

Given log 2 = 0.3010, log 3 = 0.4771. Find the value of a) log 12 b) log 5.

Answers

a) log 12 
= log3*4
= log3 + log4
=log3 + log 2^2
=log3 + 2log2
= 0.4771 +2*0.3010
=1.0791

b) log5
= log10/2
= log10 - log2
=1 - 0.3110
= 0.6989
a)\ \ \ log12=log(4\cdot3)=log4+log3=log2^2+log3=2log2+log3\approx\n\n\approx2\cdot0.3010+0.4771=0.602+0.4771=1.0791\n\n2)\ \ \ log5=log (10)/(2) =log10-log2\approx1-0.3010=0.699

Part A: Solve −np − 90 < 30 for n. Show your work. Part B: Solve 5m − 3h = 45 for h. Show your work.

Answers

A.well to start you have to do  np + 90 >- 30
np > - 20
n > - 120/p

and for b. 
5m - 45 = 3h
so h = 5m/3 - 15

Janet drove 300 miles in 4.5 hours. Write an equation to find the rate at which she was traveling.

Answers

If you would like to know the rate at which she was travelling, you can calculate this using the following steps:

300 miles ... 4.5 hours
x miles = ? ... 1 hour

300 * 1 = 4.5 * x
300 = 4.5 * x
x = 300 / 4.5
x = 200 / 3
x = 66.67 miles per hour

Janet was travelling at the rate 66.67 miles per hour. 

Can someone explain the steps to solve this so I can graph it?
y = 2x + 3?

Answers

y =mx+b   y=2x+3
m is the slope 
b is your yintercept [write it as a point   (0,b) ]
Step 1: Plot the y intercept (0,3)
Step 2: Trace slope from yintercept ... go 2 place up them 1 place right you should end up on the point (5,1)
Step 3: Draw a line through the two points

Bet ill make u a brilliest if u solve it correctly / pls dont post nontification like" report people who ask for joining meet."A car left Kamloops at 9:00 AM and arrived at Vancouver, 340 km away, at 1:15 PM the same day. What was the average speed of the car in km per hour?

Hint: The average speed in km/h is the total distance in km divided by the total time in hours.

Reminder: Calculators are not allowed.

65 km/h
70 km/h
75 km/h
80 km/h
85 km/h
90 km/h
95 km/h

Answers

Answer:

80 km/h

Step-by-step explanation:

From 9:00 am to 1:15 is 4.25 hours.

340 / 4.25 is the average speed.

Without a calculator, we can use long division.
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