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Debora [2.8K]
4 years ago
5

Circle the equation with roots 4 and -8

Mathematics
1 answer:
Arisa [49]4 years ago
4 0
Wih roots r1 and r2
the factors are (x-r1)(x-r2)

roots 4 and -8
(x-4)(x-(-8))
(x-4)(x+8)
x^2+4x-32 is the equation


either (x-4)(x+8) or x^2+4x-32
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.?????????????????????????
weqwewe [10]

Step-by-step explanation:

To divide fractions, we have to focus on 4 steps, which are:

1: Change to improper fractions

2: Keep the same

3: Change the sign to multiplication

4: Flip the 2nd fraction in the equation

So, step one:

1/3 / 4/5

Step two:

1/3 / 4/5

Step three:

1/3 x 4/5

Step four:

1/3 x 4/5 = (4 / 15)

4/15 is in simplist for.

Therefore, your answer is 4/15.

Best of Luck to you!!

4 0
3 years ago
What makes an equation false?
stellarik [79]

Answer:

When ever you can’t solve it it considered false like 4 - 12

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please solve, answer choices included.
qaws [65]
4. To solve this problem, we divide the two expressions step by step:

\frac{x+2}{x-1}* \frac{x^{2}+4x-5 }{x+4}
Here we have inverted the second term since division is just multiplying the inverse of the term.

\frac{x+2}{x-1}* \frac{(x+5)(x-1)}{x+4}
In this step we factor out the quadratic equation.


\frac{x+2}{1}* \frac{(x+5)}{x+4}
Then, we cancel out the like term which is x-1.

We then solve for the final combined expression:
\frac{(x+2)(x+5)}{(x+4)}

For the restrictions, we just need to prevent the denominators of the two original terms to reach zero since this would make the expression undefined:

x-1\neq0
x+5\neq0
x+4\neq0

Therefore, x should not be equal to 1, -5, or -4.

Comparing these to the choices, we can tell the correct answer.

ANSWER: \frac{(x+2)(x+5)}{(x+4)}; x\neq1,-4,-5

5. To get the ratio of the volume of the candle to its surface area, we simply divide the two terms with the volume on the numerator and the surface area on the denominator:

\frac{ \frac{1}{3} \pi  r^{2}h }{ \pi  r^{2}+ \pi r \sqrt{ r^{2}  +h^{2} }  }

We can simplify this expression by factoring out the denominator and cancelling like terms.

\frac{ \frac{1}{3} \pi r^{2}h }{ \pi r(r+ \sqrt{ r^{2} +h^{2} } )}
\frac{ rh }{ 3(r+ \sqrt{ r^{2} +h^{2} } )}
\frac{ rh }{ 3r+ 3\sqrt{ r^{2} +h^{2} } }

We then rationalize the denominator:

\frac{rh}{3r+3 \sqrt{ r^{2} + h^{2} }}  * \frac{3r-3 \sqrt{ r^{2} + h^{2} }}{3r-3 \sqrt{ r^{2} + h^{2} }}
\frac{rh(3r-3 \sqrt{ r^{2} + h^{2} })}{(3r)^{2}-(3 \sqrt{ r^{2} + h^{2} })^{2}}}=\frac{3 r^{2}h -3rh \sqrt{ r^{2} + h^{2} }}{9r^{2} -9 (r^{2} + h^{2} )}=\frac{3rh(r -\sqrt{ r^{2} + h^{2} })}{9[r^{2} -(r^{2} + h^{2} )]}=\frac{rh(r -\sqrt{ r^{2} + h^{2} })}{3[r^{2} -(r^{2} + h^{2} )]}

Since the height is equal to the length of the radius, we can replace h with r and further simplify the expression:

\frac{r*r(r -\sqrt{ r^{2} + r^{2} })}{3[r^{2} -(r^{2} + r^{2} )]}=\frac{ r^{2} (r -\sqrt{2 r^{2} })}{3[r^{2} -(2r^{2} )]}=\frac{ r^{2} (r -r\sqrt{2 })}{-3r^{2} }=\frac{r -r\sqrt{2 }}{-3 }=\frac{r(1 -\sqrt{2 })}{-3 }

By examining the choices, we can see one option similar to the answer.

ANSWER: \frac{r(1 -\sqrt{2 })}{-3 }
8 0
4 years ago
What is the standard form of the equation of the circle with center (-2, -3) and a radius of
lorasvet [3.4K]
It’s the last one. (X+2)2+(x+372-1/2)
7 0
3 years ago
Point a(4, 2) is translated according to the rule (x, y) right arrow (x + 1, y – 5) and then reflected across the y-axis.
Alona [7]

A: Quadrant 1

B: The coordinates are 5, -3, quadrant 2

C: -5, -3, in quadrant 3

6 0
3 years ago
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