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Nostrana [21]
3 years ago
8

Given h(x) = 2x – 3, find h(5)

Mathematics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

h(5) = 7

Step-by-step explanation:

Step 1: Define

h(x) = 2x - 3

h(5) is x = 5

Step 2: Substitute and Evaluate

h(5) = 2(5) - 3

h(5) = 10 - 3

h(5) = 7

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Factor 9x2 - 4y2 Group of answer choices
serious [3.7K]

Answer:

(3x-2y)(3x+2y)

Step-by-step explanation:

Proof:

(3x-2y)(3x+2y)-use FOIL method (First, Outer, Inner, Last)

3x*3x+2y3x-2y3x-2y*2y

9x^2+0-4y^2

<u>9x^2-4y^2</u>

The answer is (3x-2y)(3x+2y).

7 0
3 years ago
PLS HELP I WILL GIVE BRAINLIEST
drek231 [11]

Answer:

(0, -10)

Step-by-step explanation:

If you add both together from x - y, each add up to their own value. For example, -5, and -9 would be -14. It's quite simple. I hope this helps. I'm sorry if this is not what you are looking for.

4 0
3 years ago
4 7/8 + (-6 1/4) how do I find common denominators?
xxTIMURxx [149]
I hope this helps you

4 0
3 years ago
I have a F in this class may someone help me..?
Bess [88]

Answer:

x = -10

Step-by-step explanation:

9 x -10 = -90

5 0
3 years ago
Read 2 more answers
What are the roots of the polynomial equation x^3-5x+5=2x^2-5? Use a graphing calculator and a system of equations. Round nonint
leonid [27]

The right answer is c. –2.24, 2, 2.24


This question needs to be solved in two ways. First, using a graphing calculator. Next, using a system of equations.


1. Using a graphing calculator.


We have the following polynomial equation:

x^3-5x+5=2x^2-5


By ordering this equation we have:

x^3-2x^2-5x+10=0


So, we can say that this equation comes from a function given by:

f(x)=x^3-2x^2-5x+10


Thus, by plotting this function, we have that the graph of this function is indicated in Figure 1. By zooming, we can see, in Figure 2, that the roots of the polynomial equation are the x-intercepts of f(x) which are:


x_{1}=-2.236 \\ \\ x_{2}=2 \\ \\ x_{3}=2.236


Finally, rounding noninteger roots to the nearest hundredth we have:


\boxed{Root_{1}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=2.24}


2. Using a system of equations.


The ordered equation is:

x^3-2x^2-5x+10=0


By arranging to factor out we have:

x^3-5x-2x^2+10=0


Then, by factoring:

x(x^2-5)-2(x^2-5)=0


Term (x^2-5) is a common factor, thus:


(x-2)(x^2-5)=0 \\ \\ (x-2)(x-\sqrt{5})(x+\sqrt{5})=0 \\ \\ Finally: \\ \\ \boxed{Root_{1}=-\sqrt{5}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=\sqrt{5}=2.24}

6 0
3 years ago
Read 2 more answers
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