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Andrew [12]
3 years ago
13

If x= -4, find the value of x²-5+6.​

Mathematics
2 answers:
sveta [45]3 years ago
8 0

Step-by-step explanation:

If x= -4, find the value of x²-5x+6=(-4)²-5×-4+6

=16+20+6=42

umka2103 [35]3 years ago
7 0

Answer:

x^2_5+6

=-4^2-5+6

=+16-5+6

=+11+6

=17

Step-by-step explanation:

then the answer is 17

I hope it helps you c:

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Please help me......
likoan [24]

Answer:

\frac{3}{8}

Step-by-step explanation:

A=6*\frac{1}{4} ^2\\\\A=6*\frac{1}{16} \\\\A=\frac{6}{16} \\\\A=\frac{3}{8}

7 0
3 years ago
Read 2 more answers
Find the slope of the line shown on the graph to the right
damaskus [11]

Answer:

1

Step-by-step explanation:

We need two points to find the slope

( -1,0) and ( 2,3)

We can use the slope formula

m = ( y2-y1)/(x2-x1)

   = ( 3-0)/(2 - -1)

  = (3-0) / ( 2+1)

  = 3/3

  = 1

The slope is 1

6 0
3 years ago
What is the correct answer?
vovikov84 [41]

Answer:

84 units2

Step-by-step explanation:

First, you multiply 24 *7 = 168

Second, you divide 168 / 2 = 84

Lastly, have a great day!

5 0
3 years ago
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
5 0
3 years ago
Solve for AB<br><br><br> a: 3.6<br> b: 15<br> c: 4<br> d: 10
IrinaK [193]
A because 9 divided by 15 equals 0.6. Then I multiply 0.6 times 6 which equals 3.6. I didn’t 9 divided by 15 because they are congruent.
4 0
3 years ago
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