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2 years ago

What type of angle is this?

Mathematics

2 answers:

alexgriva [62]2 years ago

6 0

Answer:

Obtuse Angle

Step-by-step explanation:

It is greater than 90˚

Ymorist [56]2 years ago

5 0

The answer is “obtuse angle”

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A pair of snow boots at an equipment store in Big Bear that originally cost $80 is on sale for $56. What is the rate of discount

8_murik_8 [283]

Divide the sale price by the original price:

56 / 89 = 0.7

Multiply by 100:

0.7 x 100 = 70%

The sale price is 70% of the original price, so the discount would be 30% (100-70= 30)

4 0

3 years ago

This is correct? I need help please :(

Leto [7]

Answer:

well, we only no that

2l + 2w is 80

(lengths and widths on all 4 sides

and that w * l is A

so the function should be w times what's left for l, but also expressed by something with w

2l + 2w = 80 | -2w

2l = 80 -2w | devide by 2

l = 80 - w

now we can substitute l in

A = w * l

so that we only need w's

A(w) = w * (80-w)

for any w it will give us the area, makes only sense for 0<w<80

8 0

2 years ago

Read 2 more answers

What is the domain of the graph?

maksim [4K]

Domain:
-4 ≤ x ≤ 3
or in different form:
x ∈ [-4; 3]

4 0

3 years ago

Subject : Maths Level : High school Topic : Surds Points : 72

Mila [183]

Answer:

a = 5

Step-by-step explanation:

Simplify:

√5(√8 + √18) =
√5*8 + √5*18 =
√40 + √90 =
√4*10 + √9*10 =
2√10 + 3√10 =
(2 + 3)√10 =
5√10

The value of a:

a√10 = 5√10
a = 5

6 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago

What type of angle is this?​

What type of angle is this?