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3 years ago

Can someone please help me solve this?

Mathematics

1 answer:

maw [93]3 years ago

4 0

-1, -2, 2.
Those are the zeros
Hope this helps

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3. Jim invests $3500, compounded annually at 12% for 3 years. What is the total

irga5000 [103]

Answer:

None of the above, it should be $4760

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 12%/100 = 0.12 per year,

then, solving our equation

I = 3500 × 0.12 × 3 = 1260

I = $ 1,260.00

The simple interest accumulated

on a principal of $ 3,500.00

at a rate of 12% per year

for 3 years is $ 1,260.00

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2 years ago

If x and y are real numbers such that x > 0 and y < 0, then

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Answer:

Step-by-step explanation:

The third one

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2 years ago

If f(x)=2x+5 and g(x)=x²+5, evaluate the functions:

Ostrovityanka [42]

Answer:

f(-3) = -1

g(-3) = 14

Step-by-step explanation:

plug -3 for x to get the answer

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3 years ago

Read 2 more answers

Semicircles

Studentka2010 [4]

Answer:

The area of the shaded portion of the figure is $9.1\ cm^2$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The shaded area is equal to the area of the square less the area not shaded.

There are 4 "not shaded" regions.

step 1

Find the area of square ABCD

The area of square is equal to

$A=b^2$

where

b is the length side of the square

we have

$b=4\ cm$

substitute

$A=4^2=16\ cm^2$

step 2

We can find the area of 2 "not shaded" regions by calculating the area of the square less two semi-circles (one circle):

The area of circle is equal to

$A=\pi r^{2}$

The diameter of the circle is equal to the length side of the square

$r=\frac{b}{2}=\frac{4}{2}=2\ cm$ ---> radius is half the diameter

substitute

$A=\pi (2)^{2}$

$A=4\pi\ cm^2$

Therefore, the area of 2 "not-shaded" regions is:

$A=(16-4\pi) \ cm^2$

and the area of 4 "not-shaded" regions is:

$A=2(16-4\pi)=(32-8\pi)\ cm^2$

step 3

Find the area of the shaded region

Remember that the area of the shaded region is the area of the square less 4 "not shaded" regions:

$A=16-(32-8\pi)=(8\pi-16)\ cm^2$

---> exact value

assume

$\pi =3.14$

substitute

$A=(8(3.14)-16)=9.1\ cm^2$

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3 years ago

Factor the expression x2 − 8x + 16

Anastaziya [24]

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