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2 years ago

I RLLY NEED HELP PLS!!

Mathematics

1 answer:

lara [203]2 years ago

5 0

Answer: $(x+2)^2+3\\\\$

a = 2 and b = 3

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Explanation:

Let's expand out $(x+a)^2+b$ to get the following:

$(x+a)^2+b\\\\(x+a)(x+a)+b\\\\x(x+a)+a(x+a)+b\\\\x^2+ax+ax+a^2+b\\\\x^2+2ax+a^2+b\\\\$

The x term here is 2ax

Compare this to the x term of $x^2+4x+7$ and we see that

$2ax = 4x\\\\2a = 4\\\\a = 4/2\\\\a = 2\\\\$

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The constant term of $x^2+2ax+a^2+b\\\\$ is the $a^2+b$ portion since it doesn't have the variable x attached to it.

Compare this with the 7 of $x^2+4x+7$ which is also the constant.

Equate the two items, plug in a = 2 and solve for b.

$a^2+b = 7\\\\2^2+b = 7\\\\4+b = 7\\\\b = 7-4\\\\b = 3\\\\$

Therefore,

$x^2+4x+7 = (x+a)^2+b\\\\x^2+4x+7 = (x+2)^2+3\\\\$

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Andrew [12]

Answer:

40%

Step-by-step explanation:

If there are 1200 students in the school and 720 of them are boys that means 480 of them are girls.

I like to do cross multiply and divide so the equation for that would be 480/120 = ?/100

if you cross and multiply 480 by 100 you will get 48000 then if you divide that by 1200 you will get 40 as your missing value.

Im basically just saying 480/1200 = 40%

If you need further help there are plenty of you tube videos on how to cross multiply and divide

Hope this helps

- Please mark brainliest :)

8 0

2 years ago

Let f(x) = (x − 3)−2. Find all values of c in (1, 7) such that f(7) − f(1) = f '(c)(7 − 1). (Enter your answers as a comma-separ

Sidana [21]

Answer:

This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3

Step-by-step explanation:

The given function is

$f(x)=(x-3)^{-2}$

When we differentiate this function with respect to x, we get;

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

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)

This implies that;

$0.06-0.25=-\frac{2}{(c-3)^3} (6)$

$-0.19=-\frac{12}{(c-3)^3}$

$(c-3)^3=\frac{-12}{-0.19}$

$(c-3)^3=63.15789$

$c-3=\sqrt[3]{63.15789}$

$c=3+\sqrt[3]{63.15789}$

$c=6.98$

If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).

But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.

3 0

3 years ago

In a certain Algebra 2 class of 29 students, 7 of them play basketball and 14 of them

Mariulka [41]

Answer:

<em>Two possible answers below</em>

Step-by-step explanation:

<u>Probability and Sets</u>

We are given two sets: Students that play basketball and students that play baseball.

It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.

This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:

$\displaystyle P=\frac{19}{29}$

P = 0.66

Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:

We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.

Thus, there 19-2=17 students who play only one of the sports. The probability is:

$\displaystyle P=\frac{17}{29}$