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ivolga24 [154]
3 years ago
10

25 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!PLEASE HELP ME QUICK !!!!!!!!!!!!!!!!!!!!! Will Give Brainliest!!!!!!!Cara and Cody have t

here first job. They need to make 9 quarts of violet paint. Cody has figured out that violet paint is made by mixing 1/3 red with 2/3 blue paint.
How much red paint will they need?

How much blue paint?

Explain how you figured it out.

Is this a proportional relationship?

If so, model this using two of the three ways that proportional relationships are represented ( table, graph or equation)

Indicate what the constant of proportionality is in this problem.
Mathematics
2 answers:
maria [59]3 years ago
3 0

Answer:

a 3 quarts

b 6 quarts

c thats in the step by step

d yes

Step-by-step explanation:

c since they have to make 9 quarts of violet paint if you add the number of red and blue paint mixed together you get 1 quart meaning you have to multiply both by nine and if you simply you get what i have above

This is a proportional relationship because you decide whether two quantities are in a proportional relationship, by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.

-hope this helps :D

natita [175]3 years ago
3 0
A) 3
B) 6
C)for red you have to find 1/3 of the 9 quarts of violet paint they are going to make.
For blue you have to find 2/3 of the 9 quarts of violet paint they are going to make.
D) yes
Soz for not doin the rest.
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a^2/3

Step-by-step explanation:

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Simplify the expression.<br> 28 + 4<br> 2<br> A. 9<br> B. 11<br> C. 16<br> D. 22
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C) 16

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Find all solutions to the following quadratic equations, and write each equation in factored form.
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Answer:

(a) The solutions are: x=5i,\:x=-5i

(b) The solutions are: x=3i,\:x=-3i

(c) The solutions are: x=i-2,\:x=-i-2

(d) The solutions are: x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}

(e) The solutions are: x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i

(f) The solutions are: x=1

(g) The solutions are: x=0,\:x=1,\:x=-2

(h) The solutions are: x=2,\:x=2i,\:x=-2i

Step-by-step explanation:

To find the solutions of these quadratic equations you must:

(a) For x^2+25=0

\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\x^2+25-25=0-25

\mathrm{Simplify}\\x^2=-25

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-25},\:x=-\sqrt{-25}

\mathrm{Simplify}\:\sqrt{-25}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}\\\\\sqrt{-25}=\sqrt{-1}\sqrt{25}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\\sqrt{-25}=\sqrt{25}i\\\\\sqrt{-25}=5i

-\sqrt{-25}=-5i

The solutions are: x=5i,\:x=-5i

(b) For -x^2-16=-7

-x^2-16+16=-7+16\\-x^2=9\\\frac{-x^2}{-1}=\frac{9}{-1}\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{-9},\:x=-\sqrt{-9}

The solutions are: x=3i,\:x=-3i

(c) For \left(x+2\right)^2+1=0

\left(x+2\right)^2+1-1=0-1\\\left(x+2\right)^2=-1\\\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x+2=\sqrt{-1}\\x+2=i\\x=i-2\\\\x+2=-\sqrt{-1}\\x+2=-i\\x=-i-2

The solutions are: x=i-2,\:x=-i-2

(d) For \left(x+2\right)^2=x

\mathrm{Expand\:}\left(x+2\right)^2= x^2+4x+4

x^2+4x+4=x\\x^2+4x+4-x=x-x\\x^2+3x+4=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are:

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=3,\:c=4:\quad x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}

x_1=\frac{-3+\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}+i\frac{\sqrt{7}}{2}\\\\x_2=\frac{-3-\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}-i\frac{\sqrt{7}}{2}

The solutions are: x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}

(e) For \left(x^2+1\right)^2+2\left(x^2+1\right)-8=0

\left(x^2+1\right)^2= x^4+2x^2+1\\\\2\left(x^2+1\right)= 2x^2+2\\\\x^4+2x^2+1+2x^2+2-8\\x^4+4x^2-5

\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4\\u^2+4u-5=0\\\\\mathrm{Solve\:with\:the\:quadratic\:equation}\:u^2+4u-5=0

u_1=\frac{-4+\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad 1\\\\u_2=\frac{-4-\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad -5

\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x\\\\\mathrm{Solve\:}\:x^2=1=\quad x=1,\:x=-1\\\\\mathrm{Solve\:}\:x^2=-5=\quad x=\sqrt{5}i,\:x=-\sqrt{5}i

The solutions are: x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i

(f) For \left(2x-1\right)^2=\left(x+1\right)^2-3

\left(2x-1\right)^2=\quad 4x^2-4x+1\\\left(x+1\right)^2-3=\quad x^2+2x-2\\\\4x^2-4x+1=x^2+2x-2\\4x^2-4x+1+2=x^2+2x-2+2\\4x^2-4x+3=x^2+2x\\4x^2-4x+3-2x=x^2+2x-2x\\4x^2-6x+3=x^2\\4x^2-6x+3-x^2=x^2-x^2\\3x^2-6x+3=0

\mathrm{For\:}\quad a=3,\:b=-6,\:c=3:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:3}}{2\cdot \:3}\\\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:3}\\x=\frac{-\left(-6\right)}{2\cdot \:3}\\x=1

The solutions are: x=1

(g) For x^3+x^2-2x=0

x^3+x^2-2x=x\left(x^2+x-2\right)\\\\x^2+x-2:\quad \left(x-1\right)\left(x+2\right)\\\\x^3+x^2-2x=x\left(x-1\right)\left(x+2\right)=0

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

x=0\\x-1=0:\quad x=1\\x+2=0:\quad x=-2

The solutions are: x=0,\:x=1,\:x=-2

(h) For x^3-2x^2+4x-8=0

x^3-2x^2+4x-8=\left(x^3-2x^2\right)+\left(4x-8\right)\\x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\x^3-2x^2+4x-8=\left(x-2\right)\left(x^2+4\right)

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

x-2=0:\quad x=2\\x^2+4=0:\quad x=2i,\:x=-2i

The solutions are: x=2,\:x=2i,\:x=-2i

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