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BlackZzzverrR [31]

4 years ago

6

Evaluate the expressions.StotisEddieS?LILLESTliteitMATURE

1 answer:

denis23 [38]4 years ago

4 0

Answer:

- 1, and - 1

the answer would be -2

Step-by-step explanation:

any number of the zero power is one.

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What is this can someone help

koban [17]

⏫

$\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}$

see this attachment ☝

6 0

3 years ago

If x- 2 is a factor of x^2 -bx+ b, where b is a constant, what is the value of b

KATRIN_1 [288]

$\bf x^2-\underline{b} x+b\implies \stackrel{given}{(x-2)}~(x-y)\impliedby \textit{now let's do some FOIL}\\\\
-------------------------------\\\\
\begin{cases}
-yx-2x=-\underline{b}x\implies y+2=\underline{b}\\
(-2)(-y)=b\implies 2y=b
\end{cases}
\\\\\\
\underline{b}=b\implies y+2=2y\implies 2=y$

what is b? from the first equation in the system, y + 2 = b.

4 0

3 years ago

The $25.00 shirt costs the store $20.00. Find the markup rate.

pochemuha

Answer:

25%

Step-by-step explanation:

3 0

3 years ago

Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.

NISA [10]

Answer:

$\frac{\sqrt{\pi}}{4}$

Step-by-step explanation:

You are going to integrate the following function:

$g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}$ (1)

furthermore, you know that:

$\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}$

lets call to this integral, the integral Io.

for a general form of I you have In:

$I_n=\int_0^{\infty}x^ne^{-ax^2}dx$

furthermore you use the fact that:

$I_n=-\frac{\partial I_{n-2}}{\partial a}$

by using this last expression in an iterative way you obtain the following:

$\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}}$ (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

$\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}$

5 0

3 years ago

Read 2 more answers

Use the quadratic formula to solve the equation.–x2 + 5x = 3

Karo-lina-s [1.5K]

Given the equation - x² + 5x = 3, which can be rewritten as:

- x² + 5x - 3 = 0

where a = -1, b = 5 and c = -3.

Quadratic formula:

$\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{2a}$

Now, we just replace the values of a, b and c on the equation above.

$\frac{-5\text{ }\pm\text{ }\sqrt[]{5^2\text{ - 4(-1)(3)}}}{2(-1)}$

=

$\frac{5}{2}\text{ }\pm\text{ }\frac{\sqrt[]{13}}{2}$

4 0

1 year ago