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Anastaziya [24]
3 years ago
7

Which expression represents 75 divided by an unknown number? A. 75 – r B. 75 · r C. 75 ÷ r D. 75 + r

Mathematics
2 answers:
Harman [31]3 years ago
7 0
1. c 2. d and 3,a hope i helped

faltersainse [42]3 years ago
6 0

I think this is right

1.c

2.a

3.a

4.d

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What is 4.09 less than 5.6 ??
Gnesinka [82]

Answer: -1.51, or it could be 4.09 < 5.6 (Im sorry if I get it wrong I just wanted to help)

Step-by-step explanation:

3 0
2 years ago
Roselyn is a college student with two part-time jobs. She works a few hours a day at the college laboratory, which pays her at a
VladimirAG [237]

Answer:

15 hours.

Step-by-step explanation:

We first need to see how much money Rose would earn if she sold the 10 pendents, so we will multiply 4×10.

4×10=40

Now we know, she already has 40 dollars of the 160 dollars she needs.

To find out how much she needs left, we will subtract 40 from 160.

160-40=120

She needs a total of 120 more dollars from working.

To find out how many hours she needs to work, we will divide 120 by 8.

120÷8=15.

In conclusion, she needs to work for at least15 hours to receive enough money.Please help me by marking this brainliest, and giving a thanks!

8 0
3 years ago
If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v
pochemuha

Answer:

V'(t) = -250(1 - \frac{1}{40}t)

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = 5000(1 - \frac{1}{40}t )^2  , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = 2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )

V'(t) = 2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )

When we simplify the above, we get

V'(t) = -250(1 - \frac{1}{40}t)

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0
3 years ago
Fill in the blanks:
muminat
6x + y = 1 , y, y, -6x + 1 or 1 - 6x
7 0
3 years ago
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
3 years ago
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