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Alenkinab [10]
2 years ago
5

5/7 + 3/8?

Mathematics
1 answer:
Rudiy272 years ago
7 0

I don't know if anyone know say him answer

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Given that p= x-y <br> Find p when :<br> X=-12. And y=-19<br><br> What is p?
Mazyrski [523]

Answer:

p=7

Step-by-step explanation:

p=-12-(-19)

p=-12+19

p=7

7 0
2 years ago
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A random number generator app on your phone generates two numbers between 0 and 8. What is the probability that the first number
aev [14]

Answer:

1/27

Step-by-step explanation:

There are 9 numbers between 0 and 8.  3 of them are less than 3 (0, 1, and 2), and 1 of them is 6.

The probability is therefore (3/9) (1/9) = 1/27.

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What is the sum of 6 ×10^3 and3×10^2​​?
Andrews [41]

Answer:

6,300

Step-by-step explanation:

6000+300

expressed as standard form^

7 0
2 years ago
Which is closet to 1,0, or 1 half
Alekssandra [29.7K]

36/40 would be close to 1

8 0
3 years ago
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Will mark brainliest!!!plz helppp
muminat

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If f(x)=x^2-6x+3, then f(x-2)=(x-2)^2-6(x-2)+3.

Let's simplify that.

Distribute with -6(x-2):

f(x-2)=(x-2)^2-6x+12+3

Combine the end like terms 12+3:

f(x-2)=(x-2)^2-6x+15

Use (x-b)^2=x^2-2bx+b^2 identity for (x-2)^2:

f(x-2)=x^2-4x+4-6x+15

Combine like terms -4x-6x and 4+15:

f(x-2)=x^2-10x+19

We are given g(x)=f(x-2).

So we have that g(x)=x^2-10x+19.

The vertex happens at x=\frac{-b}{2a}.

Compare x^2-10x+19 to ax^2+bx+c to determine a,b,\text{ and } c.

a=1

b=-10

c=19

Let's plug it in.

\frac{-b}{2a}

\frac{-(-10)}{2(1)}

\frac{10}{2}

5

So the x- coordinate is 5.

Let's find the corresponding y- coordinate by evaluating our expression named g at x=5:

5^2-10(5)+19

25-50+19

-25+19

-6

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is a(x-h)^2+k where the vertex is (h,k).

Let's put f into this form.

We are given f(x)=x^2-6x+3.

We will need to complete the square.

I like to use the identity x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2.

So If you add something in, you will have to take it out (and vice versa).

x^2-6x+3

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

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

(x+-3)^2+3-9

(x-3)^2+-6

So we have in vertex form f is:

f(x)=(x-3)^2+-6.

The vertex is (3,-6).

So if we are dealing with the function g(x)=f(x-2).

This means we are going to move the vertex of f right 2 units to figure out the vertex of g which puts us at (3+2,-6)=(5,-6).

The y- coordinate was not effected here because we were only moving horizontally not up/down.

3 0
2 years ago
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