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harina [27]
3 years ago
12

Solve this equation by using the substitution method. Make sure you show ALL your work.

Mathematics
1 answer:
djverab [1.8K]3 years ago
4 0

Step-by-step explanation:

substitute for y

x+3+4 = -6x

x+ 7 = -6x

7 = -7x

-1 = x

y = -1+3

y = 2

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A sculptor is selling her work at an exhibit and hopes to earn at least $31,000 in revenue. Small pieces go for $780 and large p
Sauron [17]

The answer should be

780x + 1500y=31000

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3 0
3 years ago
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The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.
andrezito [222]

Answer:

Given the equation: 3x^2+10x+c =0

A quadratic equation is in the form: ax^2+bx+c = 0 where a, b ,c are the coefficient and a≠0 then the solution is given by :

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

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}

Simplify:

x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}

Also, it is given that the difference of two roots of the given equation is 4\frac{2}{3} = \frac{14}{3}

i.e,

x_1 -x_2 = \frac{14}{3}

Here,

x_1 = \frac{-10 + \sqrt{100- 12c}}{6} ,     ......[2]

x_2= \frac{-10 - \sqrt{100- 12c}}{6}       .....[3]

then;

\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}

simplify:

\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}

or

\sqrt{100- 12c} = 14

Squaring both sides we get;

100-12c = 196

Subtract 100 from both sides, we get

100-12c -100= 196-100

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve x_1 , x_2

x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}

or

x_1 = \frac{-10 + \sqrt{100- 96}}{6} or

x_1 = \frac{-10 + \sqrt{4}}{6}

Simplify:

x_1 = \frac{-4}{3}

Now, to solve for x_2 ;

x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}

or

x_2 = \frac{-10 - \sqrt{100- 96}}{6} or

x_2 = \frac{-10 - \sqrt{4}}{6}

Simplify:

x_2 = -2

therefore, the solution for the given equation is: -\frac{4}{3} and -2.


3 0
3 years ago
Need help please help me
ANTONII [103]
Megan:
x to the one third power =  x ^{1/3}
<span>x to the one twelfth power = </span>x ^{1/12}

<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>\frac{x ^{1/3}}{x ^{1/12}}
<span>
Since </span>\frac{ x^{a} }{ x^{b} } = x ^{a-b}
then \frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12}

Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4

\frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12} = x^{1/4}


Julie:
x times x to the second times x to the fifth = x * x² * x⁵

<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>\sqrt[32]{x* x^{2} * x^{5} }
<span>
Since </span>x^{a}* x^{b}= x^{a+b}
Then \sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }

Since \sqrt[n]{x^{m}} = x^{m/n} }
Then \sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}

Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
3 0
3 years ago
A bag contains 10 red marbles, 6 blue marbles, and 4 green marble.
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Answer:

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SIZIF [17.4K]

Answer:

27°

Step-by-step explanation:

arc RT = 66

1/2(120 - 66) = 27

8 0
4 years ago
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