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3 years ago

Multi-Step Literal Equations

Mathematics

1 answer:

Vesna [10]3 years ago

4 0

$\\ \sf\longmapsto t+\dfrac{1}{2}b=Y$

take t right side

$\\ \sf\longmapsto \dfrac{1}{2}b=Y-t$

$\\ \sf\longmapsto \dfrac{b}{2}=Y-t$

$\\ \sf\longmapsto b=2(Y-t)$

$\\ \sf\longmapsto b=2Y-2t$

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Solve for -5x < 2 step by step

lutik1710 [3]

Answer:

x > -2/5

Step-by-step explanation:

To solve this, you need to divide -5 on both sides to isolate x.

-5x < 2

x > -2/5

Because we divided by a negative number, the sign has to change direction.

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2 years ago

Mr. and Mrs. Smith plan to roof the cabin on

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3 years ago

How do i solve this? 5×{3×[9-(4+1)]}+20÷4×2????

Zielflug [23.3K]

Answer: 70

Step-by-step explanation:

Solve the equation in the innermost parentheses first that means solving 4+1 =5

The next step would be subtracting 9 from five which gives you 4

Next multiply 3 to 4 which gives you 12

Then multiply 5 with 12 which equals 60

Now you have the equation 60+20/4*2

Solve the division and multiplication part of the equation first because of the rule pemdas which shows multiplication comes before addition

First divide 20/4 which gives you 5 then multiply with 2 which gives you ten

After dividing and multiplying you are left with the equation 60+10

The answer is 70

4 0

2 years ago

Find the first 5 terms of 6n+2

brilliants [131]

Answer:

2, 8, 14, 20, 26

Step-by-step explanation:

At n=0

2 x 0 +2=2

At n=1

6 x 1 +2=8

At n=2

6 x 2 +2 = 14

At n=3

6 x 3 +2 = 20

At n=4

6 x 4 +2 = 26

6 0

2 years ago

Find the equation of a line passing through (-4,-3) and perpendicular to 3 x + 2 y = 14.

Reil [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$3x+2y=14\implies 2y=-3x+14\implies y=\cfrac{-3x+14}{2}\implies y = \cfrac{-3x}{2}+\cfrac{14}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so therefore

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}$

so we're really looking for the equation of a line whose slope is 2/3 and passes through (-4 , -3)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-4)}) \implies y+3=\cfrac{2}{3}(x+4) \\\\\\ y+3=\cfrac{2}{3}x+\cfrac{8}{3}\implies y=\cfrac{2}{3}x+\cfrac{8}{3}-3\implies y=\cfrac{2}{3}x-\cfrac{1}{3}$

3 0

1 year ago