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lutik1710 [3]
3 years ago
15

Solve for the value of y 29 (2y+1)

Mathematics
2 answers:
Vikentia [17]3 years ago
8 0

= 58y + 29

Step-by-step explanation:

29(2y + 1)

<h2><em>multiply</em><em> the</em><em> </em><em>parentheses</em><em> </em><em>by </em><em>2</em><em>9</em></h2>

<em>29 + 2y + 20</em>

<h2><em>calculate</em></h2>

<em>= 58y + 29</em>

<h2><em>there </em><em>for </em><em>the </em><em>equal</em><em> to</em><em> the</em><em> </em><em>solution</em><em> is</em><em> </em><em>5</em><em>8</em><em>y</em><em>+</em><em>2</em><em>9</em></h2>

<em>hope </em><em>it</em><em> helps</em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>

<em>follow</em><em> me</em><em> </em><em>and</em><em> mark</em><em> me</em><em> as</em><em> </em><em>brainlist</em><em> plss</em>

Nimfa-mama [501]3 years ago
4 0

Answer: y=-1/2 or -0.5

Step-by-step explanation:

29 (2y+1)

29(2y+1)=0

58y+29=0

58y=-29

y=-29/58

y=1/2

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Answer:

Question 6

\left(5a+2\right)\left(a+4\right)=\:5a^2+22a+8

Question 7

The width is 2x.

Step-by-step explanation:

                                           Question 6)

Expand

(5a+2)(a+4)

solving to the expand

(5a+2)(a+4)

  • \mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd

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so the expression becomes

=5aa+5a\cdot \:4+2a+2\cdot \:4

=5aa+5\cdot \:4a+2a+2\cdot \:4

=5a^2+20a+2a+8

adding similar elements: 20a+2a=22a

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Hence,

\left(5a+2\right)\left(a+4\right)=\:5a^2+22a+8

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Given

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width = Area/Length

          =\:\frac{8x^2}{4x}

          =2x                  ∵ \mathrm{Divide\:the\:numbers:}\:\frac{8}{4}=2

Therefore, the width is 2x.

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