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UNO [17]
3 years ago
13

45 points! Please help!

Mathematics
1 answer:
Tju [1.3M]3 years ago
5 0

Answer:

Sequence One

Step-by-step explanation:

S = a₁ / (1 − r) is the infinite sum of a geometric series where |r| < 1.

Sequence One: r = 1/5

Sequence Two: r = 2

Sequence Three: not geometric

Sequence Four: r = -3/2

Only Sequence One has |r| < 1.

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Alondra received weekly paychecks of $788.50 last year if 15.7% of her yearly income got withheld for federal income tax how muc
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Please help - and explain
Anni [7]

interesting

I'm going to call the function for the graph of A, A(x) and the function for the graph of B, B(x),and the function for the graph of C, C(x)

we know that when A(x)=0 (and x>0), C(x)=B(x) at that value of x

also, when B(x)=0, A(x)=C(x) at that value of x

we need to find the intersection points


a quadratic function can be written in form y=a(x-b)(x-c) where the roots or zeroes are at x=b and x=c and a is a constant. since we can only find 2 points, we will assume a=1 since in the other quadratic function, a=1 as well


let's use our 1st discovery first

when A(x)=0, (and x>0), C(x)=B(x)

find zeroes of A(x)

A(x)=x^2+2x-3

find where it equals 0

0=x^2+2x-3, factor, what 2 numbers mulitly to get -3 and add to get 2

0=(x-1)(x+3)

set equal to 0

0=x-1, x=1

0=x+3, x=-3

we are looking for positive root, so at x=1, C(x)=B(x)

meaining C(1)=B(1), 3(1)+3=B(1), 3+3=B(1), B(1)=6, that's one point



2nd thing

when B(x)=0, A(x)=C(x)

solving A(x)=C(x) for x<0 (since it's to the left of y axis)

x^2+2x-3=3x+3

x^2-x-6=0

factor, what 2 numbers multiply to get -6 and add to get -1

(x-3)(x+2)=0

set equal to 0

x-3=0, x=3

x+2=0, x=-2

we want x<0 so at x=-2

find y value, C(3)=3(-2)+3=-6+3=-3


we know that B(x) has a zero at (-3,0)

we also know that B(x) passes through the point (1,6)

if a quadratic function has zeroes at m and n, then it can be factored into form y=a(x-m)(x-n) where a is a constant, we assume a=1 because otherwise we can't find the function since we only know 2 points

we know that one zero is at -3,

y=(x-(-3))(x-n)

y=(x+3)(x-n)

subsitute the other point (1,6) to find n

6=(1+3)(1-n)

6=(4)(1-n)

6=4-4n

2=-4n

n=-1/2


so the equation is

y=(x+3)(x-(-\frac{1}{2})) or

y=(x+3)(x+\frac{1}{2})

the only problem is that in the picture, the graph of B crosses the x axis at a positive and negative value but the equation I have crosses the x axis at 2 negative values

but without further information, I can't solve for B

4 0
3 years ago
The following functions are
Rasek [7]

Answer:

ghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhStep-by-step explanation:

hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

8 0
3 years ago
I need help with this problem, i am confused and don't know what to do.​
klasskru [66]

Problems a through c are asking you what the values would be at those values.

What will the value of f(x) be at x=0? Let's look at the graph.

It seems at x=0, f(x)=0, meaning that f(0)=0

Let's do the same for the others.

At x=-3, f(-3)=2

At x=2, f(2)=-4

For the last question, it is giving us the f(x) value and asking us which x values would give us that.

At f(x)=3, x=-2.

*Note* this function is continuous to the right, meaning that there is another point where f(x)=3.

Hope this helps.

頑張って!

3 0
3 years ago
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