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hram777 [196]
3 years ago
5

How do I solve this 4 (2x-15)+40=6 (x+5)

Mathematics
2 answers:
pav-90 [236]3 years ago
7 0
You can do pemdas, parentasy, exponent, multiple,division,addition and subtraction
pochemuha3 years ago
4 0
4 (2x-15)+40=6 (x+5)
Distribute
8x-45+40=6x+30
combine -45 and 40
8x-5=6x+30
add 5 to both sides
8x=6x+35
subtract 6x from both sides
2x=35
divide by 2 on both sides
x=17.5
hope this helps!

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Someone help me please?
jeka57 [31]

The Answer

LCD = 12

Equivalent Fractions with the LCD

2/3 = 8/12

3/4 = 9/12

1/2 = 6/12

Solution:

Rewriting input as fractions if necessary:

2/3, 3/4, 1/2

For the denominators (3, 4, 2) the least common multiple (LCM) is 12.

LCM(3, 4, 2)

Therefore, the least common denominator (LCD) is 12.

Calculations to rewrite the original inputs as equivalent fractions with the LCD:

2/3 = 2/3 × 4/4 = 8/12

3/4 = 3/4 × 3/3 = 9/12

1/2 = 1/2 × 6/6 = 6/12

Hope This Was Helpful :D

6 0
3 years ago
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hodyreva [135]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Someone please help with the picture shown!!
serg [7]

Answer:

its between 45 and thirty I think its 45

Step-by-step explanation:

I did a similar problem.

5 0
3 years ago
Read 2 more answers
If A (-7,0), B(0,0), C(0,7), what type of angle is
devlian [24]

A right angle should be the correct answer.


Brainliest if it is correct!!!

8 0
3 years ago
1+cos2A/cos2A = tan2A/tanA prove LHS=RHS
son4ous [18]
RTP: \frac{1 + cos(2A)}{cos(2A)} = \frac{tan(2A)}{tan(A)}

LHS = \frac{1 + cos(2A)}{cos(2A)}
= \frac{1 + \frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}{\frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}
= \frac{\frac{1 + tan^{2}(A) + 1 - tan^{2}(A)}{1 + tan^{2}(A)}}{\frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}
= \frac{2}{1 - tan^{2}(A)}
= \frac{2}{1 - tan^{2}(A)} \cdot \frac{tan(A)}{tan(A)}
= \frac{\frac{2tan(A)}{1 - tan^{2}(A)}}{tan(A)}
= \frac{tan(2A)}{tan(A)}
= RHS, as required.
3 0
4 years ago
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