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2 years ago

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Mathematics

1 answer:

spin [16.1K]2 years ago

7 0

Answer:

For the first test the score was 89 and on the second test it was 96

Step-by-step explanation:

To find the answer we make the situation into an algebraic expression.

$x + (x - 7) = 185 \\ 2x - 7 = 185 \\ 2x = 185 + 7 \\ 2x = 192 \\ \frac{2x}{2} = \frac{192}{2}$

192 divided by 2 is 96; hence

$x = 96 \\ x + (x - 7) = 185 \\ 96 + (96 - 7) = 185 \\ 96 + 89 = 185$

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I need help with This question no links or I will report you

brilliants [131]

Answer:

Below.

Step-by-step explanation:

So let's do this Divide both sides.

equals to,

y+2>2

Cancel out equal terms.

y>0.

5 0

3 years ago

At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice

Lady bird [3.3K]

Answer: E. 120

The number of seniors were there in high school X at the beginning of the year = 120

Step-by-step explanation:

Given : At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4.

Let the number of juniors be 3x and the number of seniors be 4x.

Since , during the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X.

i.e. Number of juniors at the end of the year = 3x-10

Number of seniors at the end of the yea = 4x-2(10)=4x-20

At the end of the year, the ratio of juniors to seniors was 4 to 5.

$\Rightarrow\dfrac{3x-10}{4x-20}=\dfrac{4}{5}$

$\Rightarrow5(3x-10)=4(4x-20)$

$\Rightarrow15x-50=16x-80$

$\Rightarrow16x-15x=80-50$

$\Rightarrow x=30$

The number of seniors were there in high school X at the beginning of the year = 4(30)=120

Hence, the correct answer is E. 120 .

6 0

3 years ago

Explain why f(x) = x^2-x-6/x^2-9 is not continuous at x = 3.

iren [92.7K]

Answer:

See Explanation

Step-by-step explanation:

Given

$f(x) = \frac{x^2 - x -6}{x^2 - 9}$

Required

Why is the function not continuous at x = 3

First substitute 3 for x at the denominator

$f(x) = \frac{x^2 - x -6}{x^2 - 9}$

Factorize the numerator and the denominator

$f(x) = \frac{x^2 - 3x+2x -6}{x^2 - 3^2}$

$f(x) = \frac{x(x - 3)+2(x -3)}{(x - 3)(x+3)}$

$f(x) = \frac{(x+2)(x - 3)}{(x - 3)(x+3)}$

Divide the numerator and denominator by (x - 3)

$f(x) = \frac{x+2}{x+3}$

Substitute 3 for x

$f(3) = \frac{3+2}{3+3}$

$f(3) = \frac{5}{6}$

Because $f(x) = \frac{x^2 - x -6}{x^2 - 9}$ is defined when x = 3;

Then the function is continuous

8 0

3 years ago

It costs $8 per hour to rent a bike. Niko graphs this relationship using x for number of hours and y for total cost in dollars.

Maurinko [17]

Im pretty sure it would be (5, 40)!
the equation is y=8x :-)

7 0

3 years ago

Find the remainder for (x^10 + x^9 +...+x+1) divided by (x^2 - 1)

Assoli18 [71]

Since $x^2-1=(x-1)(x+1)$ , by the remainder theorem we have

$\dfrac{p(x)}{x-1} = q(x) + \dfrac{p(1)}{x-1} = q(x) + \dfrac{11}{x-1}$

where $p(x) = x^{10}+x^9+\cdots+x+1$ .

Then

$\dfrac{p(x)}{x^2-1} = q^*(x) + \dfrac{q(-1)}{x+1} + \dfrac{11}{x^2-1} = q^*(x) + \dfrac{q(-1)(x-1) + 11}{x^2-1}$

The only missing piece is q(x), which we can get through usual polynomial division:

$\dfrac{x^{10}+x^9+\cdots+x+1}{x-1} = \underbrace{x^9 + 2x^8 + 3x^7 + \cdots + 9x + 10}_{q(x)} + \dfrac{11}{x-1}$

so that

$q(-1) = (-1)^9 + 2(-1)^8 + \cdots + 9(-1) + 10 = 5$

Then the remainder we want is

$\dfrac{p(x)}{x^2-1} = q^*(x) + \dfrac5{x+1} + \dfrac{11}{x^2-1} = q^*(x) + \dfrac{5(x-1)+11}{x^2-1} = q^*(x) + \dfrac{\boxed{5x+6}}{x^2-1}$

3 0

3 years ago