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4 years ago

PLSSSS HELP SHOW WORK WILL CROWN BRAINLIEST 20pts!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

ankoles [38]4 years ago

7 0

The center is (0,0)
You are looking for a scale factor of 4..

Point a is found at (-3,1)
Point B (4,-3)
Point C (2,3)
Point D (-1,4)

Multiply each value by the scale factor of 4..

x:
Point A = -3 * 4 = -12
Point B = 4*4=16
Point C= 2*4=8
Point D=-1*4=-4

y:
Point A= 1*4=4
Point B= -3*4=-12
Point C= 3*4=12
Point D= 4*4=16

Your dilated points are:

A : (-12,4)
B : (16,-12)
C : (8,12)
D : (-4,16)

Hope this helps!

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Leto [7]

Answer:

$4194.39

Step-by-step explanation:

Two equal payments one after 3 months and another after 9 months respectively are used to settle a debt of $6000 due today.

We have to calculate the size of the equal payments at 6% compounded quarterly.

Let the amount be $x.

So, $(x + x) = $2x will be the total sum after interest.

Therefore, $6000 will be charged for 3 months at 6% interest quarterly and $(6000 - x) will be charged for (9 - 3) = 6 months at 6% interest quarterly.

The equation we can write is

$6000 (1 + \frac{6}{100} )^{1} + (6000 - x)(1 + \frac{6}{100} )^{2} = 2x$

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4 years ago

What is the sum of this infinite geometric series? 1/4, 1/5, 4/25, 16/125

Pepsi [2]

Answer:

Step-by-step explanation:

Given infinite geometric series

1/4, 1/5, 4/25, 16/125, ...

We see that

a = 1/4
r = 4/5

Sum of the series

S = a*1/(1 -r)
S = 1/4*1/(1 - 4/5) = 1/4*1/(1/5) = 1/4*5 = 5/4

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3 years ago

Read 2 more answers

F(x) is a function? Please help

Nonamiya [84]

False because multiple inputs can’t have the same output

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4 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$