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2 years ago

URGENT PLEASE !!!! Please answer correctly !!Will mark Brianliest !!!!!!!!!!

Mathematics

1 answer:

nirvana33 [79]2 years ago

4 0

Answer:

Step-by-step explanation:

Square Root of 85

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What is the equation of the circle whose center and radius are given.

muminat

Hello!

The equation for a circle is

$(x - h)^{2} + ( y - k)^{2} = r^{2}$

(h,k) is the center
r is the radius

Put in the values you know

$(x - 7)^{2} + ( y + 3)^{2} = 7^{2}$

Simplify

$(x - 7)^{2} + ( y + 3)^{2} = 49$

The equation is $(x - 7)^{2} + ( y + 3)^{2} = 49$

Hope this helps!

5 0

2 years ago

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3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

2 years ago

The rockets scored 15 baskets out of 21 attempts. How many attempts will it take for the rockets to score 40 baskets?

andre [41]

I may not be correct but ok pretty sure it’s 56

8 0

3 years ago

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What is the value of e to five decimal places

notsponge [240]

Using google calculator you can see that

e≈2.7 1 8 2 8 182846.....

Fifth decimal place is bold 8.

When we round up to 5th decimal place, we get

e≈2.7 1 8 2 8.

5 0

2 years ago

A/ab-bsquare + b/ab-asquare?

lord [1]

Answer:

$\dfrac{(a+b)}{ab}$

Step-by-step explanation:

The given expression is :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}$

It can be solved as follows :

$\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}$

So, the solution of the given expression is equal to $\dfrac{(a+b)}{ab}$ .

7 0

2 years ago