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3 years ago

What is the center of (x+5)^{2}+(y-4)^{2}=121?

Mathematics

1 answer:

Inessa [10]3 years ago

6 0

Answer: (-5,4)

Step-by-step explanation:

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PLZ HELP ME LLZ I WILL DO ANYTHING

____ [38]

Answer:

X=45

Step-by-step explanation:

45+3x=180

180-45=135

135/3=45

x has to be 45

x=45

4 0

3 years ago

If ABCD is congrunent to , pqrs, then AD is congrent to ?

katovenus [111]

Answer:

AD is congruent to RS

Step-by-step explanation:

we know that

If two figures are congruent, then its corresponding sides and its corresponding angles are congruent

In this problem

ABCD≅PQRS

then

<em>Corresponding angles</em>

∠A≅∠P

∠B≅∠Q

∠C≅∠R

∠D≅∠S

<em>Corresponding sides</em>

AB≅PQ

BC≅QR

CD≅RS

AD≅PS

7 0

3 years ago

A gym membership is offered in January for 14.99 per month. If you pay upfront for the year, you are given a 15% discount on the

hoa [83]

Answer:

$27 approx

Step-by-step explanation:

Given data

Membership fee= 14.99 per month

for a year the fee is 14.99*12= 179.88

discount = 15%

=15/100*179.88

=0.15*179.88

=26.982

Hence for a year, you will save $27 approx

6 0

3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago

write an equation for a line that is parallel to line that passes through the points (5,5),(-1,3) and contains the point (-6,-8)

dimaraw [331]

Since the two lines are parallel then they have the same slope which = 1/3

since the line passes through (-6 , -8)

y+8 = (1/3)(x+6)

3y + 24 = x + 6

x - 3y = 18 is the equation

8 0

3 years ago