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vazorg [7]
3 years ago
11

Two times a number is equal to six less than three times the number

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
8 0
2 times a number, x, is equal to 6 less than 3 times the number, x.

2x = 3x - 6
-1x = -6
x = 6

The number is 6.
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F(2) = -13 and f(-3) = 12
mario62 [17]

Answer:

f(x) = -5x-3

Step-by-step explanation:

f(x) = ax + b

f(2)=-13 & f(-3) =12

2a + b = -13

-3a+ b = 12

_________-

5a = -25

a = -5

b= -13+10= -3

so the formula of f(x) = -5x -3

5 0
3 years ago
Write an expression for the phrase.
vitfil [10]

Answer:

9+k-17

Step-by-step explanation:

sum of 9 and k

9+k

minus 17

9+k-17

5 0
3 years ago
Factor completely<br> x^2 - 3x - 28
lbvjy [14]

Answer:

(x-7)  (x+4)

Step-by-step explanation:

x^2 - 3x - 28

What two numbers multiply to -28 and add to -3

-7*4 = -28

-7+4 = -3

(x-7)  (x+4)

5 0
3 years ago
Read 2 more answers
HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!! CERTAIN ANSWERS AND I WILL MARK YOU AS BRAINIEST!!!! WISH YOU BEST OF LUCK HOPEFULLY ITS N
AURORKA [14]

Answer:

1)

Minimum is a 4th Degree

2)

Positive; Even

3)

x=-4, -1\text{ Odd Multiplicity}\\x=3\text{ Even Multiplicity}

Step-by-step explanation:

Part 1)

The minimum degree of our function will be 4.

Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.

So, it can be anything 1, 3, 5, 7, etc.

However, we will choose the lowest one, 1.

Next, we know that the graph <em>bounces off</em> at 3.

So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.

We choose the lowest one, 2.

Therefore, the minimum degree of our function will be 1+1+2 or 4.

Part 2)

The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.

Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.

This applies to all polynomials with even degrees.

Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.

Part 3)

This is similar to Part 1.

We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is: x=-4, -1, 3

We know that it<em> passes through</em> x=-4 \text{ and } x=-1 . So, these two factors must have an odd multiplicity.

However, since the graph <em>bounces off</em> x=3, this factor must have an even multiplicity.

And we're done!

7 0
3 years ago
Plz help me !!!!!!!!
vazorg [7]

You factor the numerator and denominator and cancel the common factors. So maybe that should help you

8 0
3 years ago
Read 2 more answers
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