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Nookie1986 [14]
3 years ago
8

Giving away another 17 points​

SAT
1 answer:
netineya [11]3 years ago
5 0
Thank youu!!! have a great day
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Answer:

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Explanation:

what do u mean?

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What is poooooop + pooooooop
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Pooop+pooooop= you

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What is a likely consequence of a mutation that prevents apoptosis
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Explanation:

If apostolic is for some reason prevented, it can lead to uncontrolled cell division and the subsequent development of a tumor.

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2 years ago
Two numbers differ by 8 if their product is 713 then the sum of the two numbers is.
const2013 [10]

The sum of the two numbers is equal to 70.

  • Let the first number be x.
  • Let the second number be y.

In this exercise, you're required to find two numbers by translating the word problem into an algebraic expression and then solving for the unknown variables (x and y).

Translating the word problem into an algebraic expression, we have;

Two numbers differ by 8:

x-y=8    ....equation 1.

The product of the two numbers is 713:

xy=713    ....equation 2.

To calculate the sum of the two numbers:

First of all, we would solve for each of the unknown variables (x and y).

From eqn. 1, we have:

x=8+y   ....equation 3.

Substituting eqn. 3 into eqn. 2, we have:

y(y+8)=713\\\\y^2+8y=713\\\\y^2+8y-713=0

Solving the quadratic equation by factorization, we have:

y^2-31y +23y-713=0\\\\y(y-31)+23(y-31)=0\\\\(y+23)(y-31)=0

y = -23 or 31.

For the value of x, when y = 31:

x=8+31

x = 39

Now, we can calculate the sum of the two numbers:

x+y=39+31=70

Read more on word problems here: brainly.com/question/13170908

4 0
2 years ago
The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}
AlladinOne [14]

Recall that √x has a domain of x ≥ 0.

So, f(x) is defined as long as

(x + 1)/(x - 1) ≥ 0

• We have equality when x = -1

• Otherwise (x + 1)/(x - 1) is positive if both x + 1 and x - 1 are positive, or both are negative:

\begin{cases}x+1>0 \implies x>-1 \\ x-1>0 \implies x>1\end{cases} \implies x > 1

\begin{cases}x+1

Then the domain of f(x) is

x > 1   or   x ≤ -1

On the other hand, g(x) is defined by two individual square root expressions with respective domains of

• x + 1 ≥ 0   ⇒   x ≥ -1

• x - 1 ≥ 0   ⇒   x ≥ 1

but note that g(1) is undefined, so we omit it from the second domain.

Then g(x) is defined so long as both x ≥ -1 *and* x > 1 are satisfied, which means its domain is

x > 1

f(x) and g(x) have different domains, so they are not the same function.

4 0
2 years ago
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