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Mashcka [7]
2 years ago
7

-4x - 8 = -(5x + 3) pleasee

Mathematics
2 answers:
Korvikt [17]2 years ago
8 0

hopefully it's correct!!

Anon25 [30]2 years ago
7 0

Answer:

x = 5

Step-by-step explanation:

-(5x + 3) --> -5x - 3

-4x - 8 = -5x - 3

-4x - 8 + 8 = -5x - 3 + 8 (add 8 to both sides)

-4x = -5x + 5

-4x + 5x = -5x + 5 + 5x (add 5 to both sides)

x = 5

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A yogurt costs 45p. How many yogurts can be bought for five pounds?
Iteru [2.4K]
1 yogurt = 45p
1p = ? yogurt 

1 ÷ 45 = 0.02

5 × 0.02 = 0.10

0.10 yogurt can be bought with 5p (pounds)
8 0
3 years ago
Adam is 9 years old and Billy is 11 years old. What will be the ratio of Adams’ age to Billy’s age in exactly one year’s time? G
son4ous [18]

Answer:

10 / 12

simplified:

5/6

since 5 is a prime number, we can't simplify this ratio any further

4 0
3 years ago
What is the product of the 2 solutions of the equation x^2+3x-21=0
Advocard [28]
X²+3x-21=0

1) we solve this square equation:
x=[-3⁺₋√(9+84)] / 2=(-3⁺₋√93)/2
We have two solutions:
x₁=(-3-√93)/2
x₂=(-3+√93)/2

2) we compute the product of the 2 solutions found.
[(-3-√93)/2][(-3+√93)/2] =(-3-√93)(-3+√93) / 4=
=(9-93)/4=-84/4=-21

Answer: the product of the 2 solutions of this equation is -21
3 0
3 years ago
How to solve an expression with variables in the exponents?
gtnhenbr [62]

Answer:

  use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

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You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
3 years ago
HELPPPPPP PLEASE I AM STRUGLING]
atroni [7]

10 is the answer

why are you struggling

so simple

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6 0
2 years ago
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