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azamat
3 years ago
13

What is the cube root of -1000p^12q^3

Mathematics
2 answers:
sergeinik [125]3 years ago
8 0
The answer would be -10p64q^1 
since -10^3(p^4)^3 (q^1)^3 is -1000p^12q^3
Alina [70]3 years ago
4 0

Answer:

(-10p^{4}q)

Step-by-step explanation:

we have

\sqrt[3]{-1000p^{12}q^{3}}

we know that

-1000p^{12}q^{3}=(-10)^{3} (p^{4})^{3}(q)^{3}=(-10p^{4}q)^{3}

substitute in the expression

\sqrt[3]{-1000p^{12}q^{3}}=\sqrt[3]{(-10p^{4}q)^{3}}=(-10p^{4}q)

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A construction crew is lengthening a road. The road started with a length of 56 miles, and the crew is adding 3 miles to the roa
Fiesta28 [93]

Answer:

Below

Step-by-step explanation:

The initial length of the road was 56. 56 is the y-intercept assuming that the graph of this function is a line.

so the equation is:

y= mx+56

m is the slope of the function wich is by how much the function grows.

By analogy, m is the distance added to the road each day.

● y= 3x+56

X is the number of days.

■■■■■■■■■■■■■■■■■■■■■■■■■■

To find the length of the road after 33 days, replace x by 33.

y= 3*33+56 = 155

So after 33 days the road is 155 miles.

5 0
3 years ago
A​ country's people consume 6.6 billion pounds of candy​ (excluding chewing​ gum) per year. Express this quantity in terms of po
german
<span>A​ country's people consume 6.6 billion pounds of candy​ (excluding chewing​ gum) per year. Population of the country is 303 million. Specific consumption, = ((6.6*10^9 pounds)/( year* 303*10^6 persons))*(year/ 12 months) =1.8 pounds/(months*persons)</span>
4 0
4 years ago
I'm completely loss I need help on the entire process to solve this problem
Alja [10]
Well if upit gives you anything else then that would be great
3 0
4 years ago
Write the fractions as fractions with a common denominator 3/4 and 1/3
JulijaS [17]

3/4 and 1/3

4x3 = 12 that will be the common denominator.

For 3/4 we first divide 12/4 = 3 and then multiply 3 x3 = 9 that's the numerator

For 1/3 we first divide 12/3 = 4 and then multiply 1x4 = 4 that's the numerator

The fractions will be rewrited as: 9/12 and 4/12

3 0
2 years ago
What are the zeros of the quadratic function f(x) = 6x2 + 12x - 7?<br>​
mina [271]

Answer:

\frac{-6+\sqrt{78} }{6} and \frac{-6-\sqrt{78} }{6}

Step-by-step explanation:

We are asked that what are the zeros of the quadratic function f(x) = 6x² +12x -7.

So, we have to find the roots of the equation, f(x) = 6x² +12x -7 =0 ...... (1)

Since the quadratic function can not be factorized, so we have to apply Sridhar Acharya's formula.

This formula gives if, ax² +bx +c =0, the the two roots of the equation are

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

Therefore, in our case 'a' being 6, 'b' being 12 and 'c' being -7, the two roots of the equation (1) will be

\frac{-12+\sqrt{12^{2}-4*6*(-7) } }{2*6},  \frac{-12-\sqrt{12^{2}-4*6*(-7) } }{2*6}

= \frac{-12+\sqrt{312} }{12},\frac{-12-\sqrt{312} }{12}

=\frac{-6+\sqrt{78} }{6} and \frac{-6-\sqrt{78} }{6}

Hence, x= \frac{-6+\sqrt{78} }{6}

and x= \frac{-6-\sqrt{78} }{6}

(Answer)

9 0
4 years ago
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