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SVEN [57.7K]
2 years ago
8

GUYS PLEASE HELP ME PLEASE HELP ME PLEASE PLEASE!!!!!

Mathematics
1 answer:
Pani-rosa [81]2 years ago
3 0

Answer:

2y-8

Step-by-step explanation:

-y-5+y+2(2y-y)-3

Do the parenthesis first

-y-5+y+2(y)-3

Multiply

-y-5+y+2y-3

Combine like terms

2y-8

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Mary's total bill for a used car is $2,700.00. If she pays an equal amount each month for an entire year, how much will she have
nasty-shy [4]

Answer:

  $225

Step-by-step explanation:

When Mary pays the cost in 12 equal payments, each is 1/12 of the total amount.

__

  1/12 × $2700 = $225

Mary will have to pay $225 each month.

4 0
1 year ago
Girardo is using the model below to solve the equation 3x+1=4x+(-4)
V125BC [204]

Answer:

x=5

Step-by-step explanation:

3x+1= 4x+(-4)

-4×+1= -4x-1 combine like terms

-1x= -5 divided by -1

x= 5

5 0
2 years ago
Simplify (9x^3-4x+10)+(x^3+10x-9)
Alex

Simplify

9x^3 - 4x + 10 + x^3 + 10x - 9

Collect like terms

(9x^3 + x^3) + (-4x + 10x) + (10 - 9)

Simplify

<u>10x^2 + 6x + 1</u>

6 0
3 years ago
Total cost of three shirts $25.99 each with sales tax of 5.5%
irakobra [83]
3 shirts at 25.99 each : 3 * 25.99 = 77.97
plus sales tax of 5.5% : 77.97 + .055(77.97) = 77.97 + 4.29 = 82.26
6 0
2 years ago
Read 2 more answers
If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

6 0
2 years ago
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