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vlada-n [284]
3 years ago
5

3(x+2y)=7y-9 when y=3

Mathematics
2 answers:
8_murik_8 [283]3 years ago
8 0
I'll do it step by step:

3(x+2(3))=7(3)-9
3(x+6)=21-9
3x+18=12

I don't know what x is, so I think that's as much as I can do. 
ANTONII [103]3 years ago
8 0
First substitute all of the y's with the value of y, like so:
3{x+(2*3)} = 7*3-9
Solve for x.
3{x+(6)} = 21-9
3(x+6)=12
x+6=4
x=(-2)
Hope this helps, please mark brainliest, I am one away from my next rank, and have an amazing day.
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Simplify 7.32a+2.1∙(2.7−18a)
tresset_1 [31]
7.32a + 2.1(2.7 -18a)
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For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10
professor190 [17]

The true statements are:

  • Tyson's expression is not equivalent to the original expression
  • The equivalent expression is:\frac{1}{2}g- \frac 1{5}

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}

Collect like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10}  -\frac{1}{10}

Evaluate the like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}

Tyler's equivalent expression is given as:

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}

Collect like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}

Evaluate the like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

3 0
2 years ago
Triangle Abc has vertices A(-2,5),B(1,0),andC(6,-2).what are the coordinates of the vertices of A’B’C for Ry-axis?
Andreyy89

Answer:

A'(2,5),B'(-1,0),C'(-6,-2)

Step-by-step explanation:

we know that

The rule of the reflection of a point across the y-axis is equal to

(x,y) -----> (-x,y)

so

Applying the rule of the reflection across the y-axis

A(-2,5) ----> A'(2,5)

B(1,0) -----> B'(-1,0)

C(6,-2) ---> C'(-6,-2)

4 0
3 years ago
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