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inna [77]
3 years ago
7

A painter leans a 25-ft ladder against a building. The base of the ladder is 7 ft from the building. To the nearest foot, how hi

gh on the building does the ladder reach?
Mathematics
1 answer:
Komok [63]3 years ago
8 0
The easiest way to solve this problem would be to use the pythagorean theorem. Plug in the information you know. 25 will have to be c because it is the hypotenuse. You can plug 7 in for a or b. It does not matter which one you plug 7 in for because you will still get the same answer. 

This is what it should look like:
7^2 + b^2 = 25^2

Now solve the equation. 
7^2 = 49
25^2 = 625

The new equation will be 49 + b^2 = 625

Now subtract 49 from 625. 

625 - 49 = 576

b^2 = 576

Now take the square root of both sides to get b by itself. 

b = 24

Your answer should be: The ladder reaches 24 ft. high on the building. 
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pychu [463]

Step-by-step explanation:

18×RM 8= RM 144

13× RM 11=RM 143

RM 144+143= RM 287

are u Malaysian too? coz I'm one hehe

8 0
3 years ago
Read 2 more answers
Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy
Margaret [11]

Answer:

Part 1) The exact solutions are

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})   and  (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Part 2) (1.79, 8.58)

Step-by-step explanation:

we have

y=x^{2} +3x ----> equation A

y=2x+5 ----> equation B

we know that

When solving the system of equations by graphing, the solution of the system is the intersection points both graphs

<em>Find the exact solutions of the system</em>

equate equation A and equation B

x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^{2} +x-5=0  

so

a=1\\b=1\\c=-5

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}

x=\frac{-1\pm\sqrt{21}} {2}

so

The solutions are

x_1=\frac{-1+\sqrt{21}} {2}

x_2=\frac{-1-\sqrt{21}} {2}

<em>Find the values of y</em>

<em>First solution</em>

For x_1=\frac{-1+\sqrt{21}} {2}

y=2(\frac{-1+\sqrt{21}} {2})+5

y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}

The first solution is the point (\frac{-1+\sqrt{21}} {2},4+\sqrt{21})

<em>Second solution</em>

For x_2=\frac{-1-\sqrt{21}} {2}

y=2(\frac{-1-\sqrt{21}} {2})+5

y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}

The second solution is the point (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Round to the nearest hundredth

<em>First solution </em>

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21}) -----> (1.79,8.58)

(\frac{-1-\sqrt{21}} {2},4-\sqrt{21}) -----> (-2.79,-0.58)

see the attached figure to better understand the problem

6 0
4 years ago
Is the answer to this question correct?
Komok [63]
It looks about right to me.
5 0
3 years ago
Read 2 more answers
Which lines are perpendicular to the line y - 1 = 1/3 (x+2)? Check all that apply.
ioda

Answer: Options 1, 3, 5

Step-by-step explanation:

Perpendicular lines have slopes that are negative reciprocals, so since the slope of the given line is 1/3, we need to find lines with a slope of -3.

  • Option 1 has a slope of -3.
  • Option 2 has a slope of 3.
  • Option 3 has a slope of -3.
  • Option 4 has a slope of 1/3.
  • If we subtract 3x from both sides, we get y=-3x+7, so option 5 has a slope of -3.
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△ABC is given with line m drawn through A parallel to BC¯¯¯¯¯¯¯¯. In the course of proving that the interior angle measures of △
aleksklad [387]

Answer:

I think a) and c is correct.

4 0
3 years ago
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