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vovikov84 [41]
3 years ago
14

Factor this expression. x2 + 9x + 8

Mathematics
1 answer:
Ket [755]3 years ago
6 0

Answer:

(x+1)(x+8) When factoring squares whose squared coefficient is one the roots must add up to the coefficient of the slope and multiply out to the intercept value.

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I will give brainliest answer
adell [148]

Answer: A=351.68 cm²

Step-by-step explanation:

To find the area of the shaded region, we would subtract the area of the circle by the area of the inner circle.

Area of Circle

A=\pi r^2

A=\pi (11)^2

A=121\pi

Area of inner (white) circle

A=\pi r^2

A=\pi (3)^2

A=9\pi

Now that we have the area to the circle and inner circle, we would subtract to find the area of the shaded region.

A=121\pi -9\pi

A=112\pi

A=351.68 cm^2

3 0
3 years ago
Please solve these for 20 points : solving formulas for a variable
Digiron [165]
6. (A/pi = r^2)
7. [(P - 2l)/2 = w]
8. [C/(2pi)= r]
9. (2A/h = b)
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6 0
3 years ago
Which statements are true based on the diagram?
s344n2d4d5 [400]

Answer:

The answer to this is a, b, d, e

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x
aalyn [17]

f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}

f must be continuous in order to be differentiable, so we need to have

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)

By its definition, f(1)=1^3=1, and

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1

\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b

so that \boxed{m+b=1}.

We want the derivative to exist at x=1, which requires that we pick an appropriate value for f'(1) so that f'(x) is also continuous. At the moment, we know

f'(x)=\begin{cases}3x^2&\text{for }x1\end{cases}{/tex]so we need to pick [tex]f'(1) such that

\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)

We have

\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3

\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m

so that \boxed{m=3} (which means we need to pick f'(1)=3) and so m+b=1\implies\boxed{b=-2}

5 0
3 years ago
What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ​ ( 3x 2 −4 2x+3 ​ )start fraction, d, divided by, d, x
stellarik [79]

Answer:

4.

Step-by-step explanation:

We are asked to find the value of expression \frac{d}{dx}(\frac{2x+3}{3x^2-4}) at x=-1.

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}

\frac{d}{dx}(\frac{2x+3}{3x^2-4})

\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}

\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}

\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}

\frac{-6x^2-18x-8}{(3x^2-4)^2}

Therefore, our required derivative is \frac{-6x^2-18x-8}{(3x^2-4)^2}.

Now, we will substitute x=-1 in our derivative to find the required value as:

\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}

\frac{-6(1)+18-8}{(3(1)-4)^2}

\frac{-6+18-8}{(3-4)^2}

\frac{4}{(-1)^2}

\frac{4}{1}

4

Therefore, the value of expression \frac{d}{dx}(\frac{2x+3}{3x^2-4}) at x=-1 is 4.

6 0
3 years ago
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