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3 years ago

1,0005.48 m2 used for 3.14pi

Mathematics

1 answer:

Valentin [98]3 years ago

6 0

Answer:

what do you mean?

Step-by-step explanation:

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ABCD IS A TRAPEZIUM WORK OUT THE SIZE OF ANGLE CDA.

zvonat [6]

Answer:

The measure of angle CDA is $27.6\°$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In the right triangle ABF

Applying Pythagoras Theorem

$A.F^{2}=6.5^{2}-6^{2}$

$A.F^{2}=6.25$

$A.F=2.5\ cm$

In the right triangle CED

$CE=6\ cm$

$ED=23-(9+2.5)=11.5\ cm$

$tan($

substitute the values

$tan($

$($

7 0

3 years ago

Solve. j+k=3j−k=7 Use the substitution method.

il63 [147K]

Answer:

the solution is (5,-2)

8 0

3 years ago

I need help please my home work not to sure how to do it

Artyom0805 [142]

1. x+5y=8
2x- 5y=1 What you do is find the number/variable that are the same( here is is the +5y and -5y. You want them to be opposites(-,+) so that they cancel each other out. if they are not opposites, then multiply one equation by a negative sign to change it. Cross out the 5y's. Now you can combine like terms-x+2x=3x, and 8+1=9, so now your equation will look like this:
3x=9
x=9/3
x=3
now take x=3 and plug it into the first equation to solve for y. x+5y=8, so 3+5y=8
5y=8-3
5y=5
y=5/5
y=1
(3,1) are your answers.
now do the same with the other problems.

4 0

3 years ago

Read 2 more answers

1 y

Nesterboy [21]

Answer:

A, B & C form the vertices of a triangle.

∠ CAB = 90°, ∠ ABC = 73° and AB = 9.4.

Calculate the length of BC rounded to 3 SF

3 0

3 years ago

Find the derivative of f(x)= (e^ax)*(cos(bx)) using chain rule

Vikentia [17]

$f(x) = e^{ax}\cos(bx)$

then by the product rule,

$f'(x) = \left(e^{ax}\right)' \cos(bx) + e^{ax}\left(\cos(bx)\right)'$

and by the chain rule,

$f'(x) = e^{ax}(ax)'\cos(bx) - e^{ax}\sin(bx)(bx)'$

which leaves us with

$f'(x) = \boxed{ae^{ax}\cos(bx) - be^{ax}\sin(bx)}$

Alternatively, if you exclusively want to use the chain rule, you can carry out logarithmic differentiation:

$\ln(f(x)) = \ln(e^{ax}\cos(bx)} = \ln(e^{ax})+\ln(\cos(bx)) = ax + \ln(\cos(bx))$

By the chain rule, differentiating both sides with respect to x gives

$\dfrac{f'(x)}{f(x)} = a + \dfrac{(\cos(bx))'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a - \dfrac{\sin(bx)(bx)'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a-b\tan(bx)$

Solve for f'(x) yields

$f'(x) = e^{ax}\cos(bx) \left(a-b\tan(bx)\right) \\\\ f'(x) = e^{ax}\left(a\cos(bx)-b\sin(bx))$

just as before.

4 0

3 years ago