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SashulF [63]
3 years ago
15

How many times greater is the area of circle B compared to the area of circle A?

Mathematics
1 answer:
givi [52]3 years ago
4 0
Your answer is the area of circle B is 2.25 times larger than circle A. 

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Two sets of three consecutive integers have a property that the product of the larger two is
skelet666 [1.2K]

Answer:

{1, 2, 3},  {3, 4, 5}

Step-by-step explanation:

Write expressions for three consecutive integers:  n, n + 1, n + 2.

Set up an equation for the verbal description: the product (mulitplication) of the two larger integers (the last two) is one less than 7 times the smallest (the first one).

(n + 1)(n + 2) = 7n - 1

Multiply (FOIL) the left side.

n^2 + 3n + 2 = 7n - 1

Subtract 7n and subtract 1 to make the right side 0.

n^2 - 4n + 3 = 0

Factor.

(n - 1)(n - 3) = 0

Set the two factors equal to 0

n - 1 = 0,  n - 3 = 0

Solve for n.

n = 1,  n = 3

One set of integers begins with 1, so it's {1, 2, 3}.

The other set begins with 3, so it's {3, 4, 5}

5 0
2 years ago
There are 158 students registered for American History classes. There are twice as many students registered in second period as
Margarita [4]

There are 28 students in first period and 56 students in sceond period and 74 students in third period

<em><u>Solution:</u></em>

Let the number of students in first period be "x"

Let the number of students in second period be "y"

Let the number of students in third period be "z"

<em><u>There are 158 students registered for American History classes.</u></em>

Therefore,

x + y + z = 158 ---------- eqn 1

<em><u>There are twice as many students registered in second period as first period</u></em>

number of students in second period = twice of number of students in first period

y = 2x ------- eqn 2

<em><u>There are 10 less than three times as many students in third period as in first period</u></em>

number of students in third period = 3 times number of students in first period - 10

z = 3x - 10 ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

x + 2x + 3x - 10 = 158

6x = 168

<h3>x = 28</h3>

<em><u>Substitute x = 28 in eqn 2</u></em>

y = 2(28)

<h3>y = 56</h3>

<em><u>Substitute x = 28 in eqn 3</u></em>

z = 3(28) - 10

z = 84 - 10

<h3>z = 74</h3>

Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period

5 0
2 years ago
ms. wilson draws a model of the factorization of a polynomial with integer factors. her model is partially complete. which equat
Oliga [24]
I'll just factor the above equation.

x² + 18x + 80

x² ⇒ x * x
80
can be:
1 x 80
2 x 40
4 x 20
5 x 16
8 x 10  Correct pair

(x+8)(x+10)
x(x+10) +8(x+10) ⇒ x² + 10x + 8x + 80 = x² + 18x + 80

x+8 = 0
x = -8

x+10 = 0
x = -10

x = -8

(-8)² + 18(-8) + 80 = 0
64 - 144 + 80 = 0
144 - 144 = 0
0 = 0

(-10)² + 18(-10) + 80 = 0
100 - 180 + 80 = 0
180 - 180 = 0
0 = 0

I think the algebra tiles will not be a good tool to use to factor the quadratic equation because the equation is not a perfect square quadratic equation.
4 0
2 years ago
Read 2 more answers
B. How was the original $20,000 investment split between the two interest rates?
Roman55 [17]

Answer:

t + f = 2000

let  represent the amount invested at three percent; let f represent the amount invested at five percent. The total of these amounts is $20,000. The total of the interest earned by each amount is $800. Of course the interest earned is the amount multiplied by the interest rate.

4 0
2 years ago
Solve for x: 12-2x=-2(y-x)
a_sh-v [17]
12 - 2x = -2(y - x)

12 - 2x = -2y - (-2x)

12 - 2x = -2y + 2x

12 - 2x - 2x = -2y + 2x - 2x

12 - 4x = -2y

12 - 12 - 4x = -2y - 12

-4x = -2y - 12

-4x/4 = -2y/4 - 12/4

-x = -0.5y - 3

-x/-1 = -0.5y/-1 - 3/-1

x = 0.5y + 3

or

x = 3 + 0.5y
3 0
2 years ago
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