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maxonik [38]
3 years ago
8

3(y + 5) - (4y - 8) = -2y + 10

Mathematics
2 answers:
liubo4ka [24]3 years ago
6 0

Answer:

y = -13

Step-by-step explanation:

3(y + 5) - (4y - 8) = -2y + 10

3y + 15 - 4y + 8 = -2y + 10

Collect like terms

3y - 4y + 2y = 10 - 15 - 8

y = -13

omeli [17]3 years ago
3 0
Y = -13!! Hope that helps
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Your friend says the solution to 4 x plus 24 x minus 2 equals 7 left parenthesis 4 x plus 2 right parenthesis is xequals16. Solv
kogti [31]

Answer:

Step-by-step explanation:

The given equation is

4x + 24x - 2 = 7(4x +2)

First step is to expand the parenthesis on the left hand side by multiplying each term inside the parenthesis by the term outside the parenthesis. It becomes

4x + 24x - 2 = 28x + 14

Collecting like terms such that the terms containing x will be on the lefthand side and constant terms will be on the right hand side, it becomes

28x - 28x = 14 + 2

0 = 16

0 cannot be equal to 16. Therefore, there is no solution.

The likely error of my friend was

28x - 28x = x

8 0
3 years ago
NEED HELP ASAP, WILL GIVE BRAINLIEST
Scrat [10]
D.........................
4 0
3 years ago
Evaluate each function following the specification.
telo118 [61]

Answer:

#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2

b. No restrictions for all real numbers

\#4 \  a. \ h(-1) = \dfrac{1}{3}, \  h(0) =0, \  h(2) =  \infty

b. Yes, <em>x ≠ 2</em>

Step-by-step explanation:

#3 The function is given as g(x) = -x² + 3

a. From the given function, by plugging in the value of 'x' in the bracket, we have;

g(-1) = -(-1)² + 3 = -1 + 3 = 2

g(-1) = 2

g(0) = -0² + 3 = 3

g(0) = 3

g(1) = -1² + 3 = -1 + 3 = 2

g(1) = 2

g(-1) = 2, g(0) = 3, and g(1) = 2

b. The given function g(x) = -x² + 3 for finding the value of <em>g</em> can take any value of <em>x</em> which is a real number

Therefore, therefore, there are no restrictions

#4 a. The given function is given as follows;

h(x) = \dfrac{x}{x - 2}

By substitution, we get;

h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}

\therefore h(-1) = \dfrac{1}{3}

h(0) = \dfrac{0}{0 - 2} = \dfrac{0}{-2} =0

\therefore h(0) =0

h(2) = \dfrac{2}{2 - 2} = \dfrac{2}{0} = \infty

\therefore h(2) =  \infty

h(-1) = \dfrac{1}{3}, \  h(0) =0, \  h(2) =  \infty

b. From the values of the function, we have that h(x) is not defined at x = 2

Therefore, there is a restriction for <em>x</em> in the function, which is <em>x ≠ 2</em>

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3 years ago
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Answer:

\frac{18p-5}{30}

Step-by-step explanation:

8 0
3 years ago
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