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lapo4ka [179]
3 years ago
13

Find slope of 12x+3y=6

Mathematics
2 answers:
lapo4ka [179]3 years ago
4 0
Y=-4x+2


Y=mx+b
12x+3y=6
3y=-12x+6
/3 /3
Y=-4x+2
sashaice [31]3 years ago
4 0

Answer:

The slope should be -4

Step-by-step explanation:

Use the formula y=mx+b to find this answer.

Sorry if I am wrong

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A toy store sold 405 dolls it sold equal number of dolls each month for 8 months and also sold a few extra dolls during the last
Simora [160]
Answer: 5 Extra dolls

Explanation: 400 divide by the 8 months equals 50 plus the 5 extra from 405
3 0
3 years ago
2 plus 2 = u btw it’s just 4
Lisa [10]

Answer:

Okay

Step-by-step explanation:

5 0
3 years ago
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Last year at Tamara made 42,789. How much did she make each month (12 months)?
zavuch27 [327]

Answer:

$3,565.75

Step-by-step explanation:

42,789 divided by 12 is 3,565.75

7 0
3 years ago
A rectangle has an area of 72 in². The length and the width of the rectangle are changed by a scale factor of 3.5. What is the a
kifflom [539]

Answer:

The area of the new rectangle is 882 in².

Step-by-step explanation:

Let l be the length and w be the width of the original rectangle,

So, the area of the original rectangle is,

A = l × w                  ( Area of a rectangle = Length × Width )

Given, A = 72 in²,

⇒ lw = 72 ------- (1),

Since, if the rectangle are changed by a scale factor of 3.5,

⇒ New length = 3.5 l,

And, new width = 3.5 w,

Thus, the area of the new rectangle = 3.5l × 3.5w

=(3.5)^2lw

=12.25\times 72  ( From equation (1) ),

= 882 in²

4 0
3 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

7 0
2 years ago
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