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gregori [183]
2 years ago
12

PLEASE ANSWER URGENT!!!!

Mathematics
2 answers:
hodyreva [135]2 years ago
8 0

the answer for number 19. is 10

the answer for number 20 is 9

Step-by-step explanation:

lara31 [8.8K]2 years ago
6 0
2, 4, 6, 8, 12
3, 6, 9, 12, 15
5, 10, 15, 20, 25

give me brainlyist?
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Check screen shot pls
IceJOKER [234]

Answer:

The last choice

Step-by-step explanation:

3 0
3 years ago
The difference between the product of 4 and a number 6
Mice21 [21]

Answer:

If we call our unknown value x, then we can say that

The "product of 4 and a number" is 4x

The "difference between" 4x and 6 is 4x - 6

Step-by-step explanation:

8 0
3 years ago
How to factor 6n2 - 6n - 12
alexandr402 [8]

We see that in these 3 terms, 6 is a common factor. So, let's factor out a 6:

6n^2-6n-12 = 6(n^2-n-2)

We can set this equal to 0 and factor by using the quadratic formula which is:

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

So, let's do just that:

6(n^2-n-2) = 0

Note that the 6 goes away if you divide both sides by 6. In this case, a = 1, b = -1, and c = -2. Let's plug that into the quadratic equation:

\frac{1 \pm \sqrt{(-1)^2-4(1)(-2)} }{2(1)} = \frac{-1 \pm \sqrt{1-(-8)} }{2}

\frac{-1 \pm \sqrt{9} }{2} = {1, -2}

So, we can write this as:

6n^2 - 6n - 12 = 6(n-1)(n+2)

Notice that the 6 comes back because it was only temporarily mad. And that the roots have opposite signs in the parentheses because to find the roots, you need to set each parentheses equal to 0 and solve for n. With that in mind, your final answer is 6(n-1)(n+2). Hope I could help you!

7 0
3 years ago
F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))
Marrrta [24]

Given:

f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6

Required:

We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.

Explanation:

Given equation is

F(x)=\sqrt[4]{f(x)g(x)}.F(x)=(f(x)g(x))^{\frac{1}{4}}F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}

Differentiate the given equation for x.

Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{  Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.

F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)

Replace x=3 in the equation.

F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}F^{\prime}(3)=\frac{-6-1}{4}F^{\prime}(3)=\frac{-7}{4}

Final answer:

F^{\prime}(3)=\frac{-7}{4}

8 0
1 year ago
Change this radical to an algebraic expression with fractional exponents.
vekshin1

Rational exponents

a^\frac{m}{n}

work like this: the numerator is the actual exponent of the base, while the denominator is the index of the root.

In other words, we have

a^\frac{m}{n}=\sqrt[n]{a^m}

So, in you case, we have

\sqrt[5]{x^3}=x^\frac{3}{5}

Assuming that the question contanis a typo. If you actually mean 5\sqrt{x^3},

then you can write it as 5x^\frac{1}{3}

7 0
3 years ago
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