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Yuki888 [10]
4 years ago
9

15-5(365x) = 1,840 What does x equal?

Mathematics
2 answers:
nirvana33 [79]4 years ago
6 0

Answer:

The answer to your question is x = -1

Step-by-step explanation:

                         15 - 5(365x) = 1840

Process

1.- Subtract 15 in both sides of the equation

                       15 - 5(365x) - 15 = 1840 -15

2.- Simplify

                            -5(365x) = 1825

3.- Divide by -5 both sides

                            -5(365x)/-5 = 1825/-5

4.- Simplify

                                365x = -365

5.- Divide by 365 both sides

                                365x/365 = -365/365

6.- Simplify

                                x = -1                                                  

vazorg [7]4 years ago
4 0

Answer:

Step-by-step explanation:

Subtract 15 in both sides of the equation

 15 - 5(365x) - 15 = 1840 -15

 -5(365x) = 1825

Divide by -5 both sides

                          

-5(365x)/-5 = 1825/-5

-365x=365

Divide by 365 both sides

 365x/365 = -365/365

x = -1          

Answer check:

substitute for (x)               

15-5(365(1))

15-5(365)

distribute -5

15-(-1825)

two negative becomes positive

15+ 1825

= 1,840

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Step-by-step explanation:

7 0
3 years ago
How many pairs of whole numbers have a sum of 40
Ahat [919]
It is 20 
39+1
38+2
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3 years ago
Please help me find n. 1/3=n+3/4
raketka [301]

Answer:

-5/12

Step-by-step explanation:

Let's solve your equation step-by-step.

1 /3  = n +  3 /4

Step 1: Flip the equation.

n +  3 /4  =  1/ 3

Step 2: Subtract 3/4 from both sides.

n+  3 /4  −  3/4   =  1/ 3  −  3 /4

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7 0
3 years ago
Select the two values of x that are roots of this equation.<br> 2x+11x+15= 0
Amiraneli [1.4K]

Answer:

The roots of the equation are x=-3 and x=-2.5

Step-by-step explanation:

<u><em>The correct quadratic equation is</em></u>

2x^2+11x+15=0

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

2x^{2} +11x+15=0  

so

a=2\\b=11\\c=15

substitute in the formula

x=\frac{-11(+/-)\sqrt{11^{2}-4(2)(15)}} {2(2)}

x=\frac{-11(+/-)\sqrt{121-120}} {4}

x=\frac{-11(+/-)\sqrt{1}} {4}

x=\frac{-11(+/-)1} {4}

x_1=\frac{-11(+)1}{4}=-2.5

x_2=\frac{-11(-)1}{4}=-3

therefore

The roots of the equation are x=-3 and x=-2.5

5 0
3 years ago
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