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oksian1 [2.3K]
3 years ago
9

Solve the system of substitution y=6x-14 y=-8x

Mathematics
1 answer:
Ivan3 years ago
8 0
<span><span><span>6x−14y</span>+<span>−6x</span></span>=<span><span>−8x</span>+<span>−6x

</span></span></span><span><span>−14y</span>=<span>−<span>14x
</span></span></span><span><span><span>
−
<span>14y/</span></span><span>−14</span></span>=<span><span>−<span>14x/</span></span><span>−14</span></span></span><span>
y=
<span>x

</span></span>
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5b= -G+ 1/5 UA Solve for a:
enot [183]

Answer:

switch sides

-g+2/5UA=5b

add g to both sides

-g+1/5UA+g=5b+g

simplify

1/5UA=5b+g

multiply both side by 5

5×1/5UA=5×5b+g

simplify

AU=25b+5g

divide both sides by U

AU/U=25b/u+5g/u =U/U=0

simplify

A=25b+5g/U

8 0
3 years ago
Seven more than one half a number as an equation
romanna [79]

Answer:

1/2x+7

Step-by-step explanation:

5 0
3 years ago
The relationship between x and y is proportional. When x is 3, y is 51. What is x when y is 204? Enter your answer in the box.​
JulsSmile [24]

Answer:

12

Step-by-step explanation:

Use the relation between 3 is to 51

eventually... you will

5 0
3 years ago
Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0
3 years ago
A ball is rolling in a circular path that has a radius of 10 inches as shown in
elena55 [62]
The complete question in the attached figure

we know that
[circumference]=2*pi*r
for r=10 in
[circumference]=2*pi*10-----> 62.8 in

if 360° (full circle) has a length of --------> 62.8 in

 54°------------------------------> X
x=54*62.8/360-----> x=9.42 in

the answer is 9.42 in

5 0
3 years ago
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