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NikAS [45]
3 years ago
5

Given h(x) = -x + 2, find h(2)​

Mathematics
1 answer:
Temka [501]3 years ago
5 0

Answer:

h(2) = 0

Step-by-step explanation:

To evaluate h(2) , substitute x = 2 into h(x), that is

h(2) = - 2 + 2 = 0

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Both of these lines share the relationship together of being parallel. 
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9. What’s the answer to this please
Lady bird [3.3K]
<h2>Answer:</h2><h3>W = 5</h3><h3>Step-by-step explanation:</h3><h3>Simplify the brackets. </h3><h3>-2x^2 + wx - 4 - x^2 - 5x - 6 = -3x^2 - 10</h3><h3>Then simplify (-2x^2 + wx - 4 - x^2 - 5x - 6) to </h3><h3>( -3x^2 + wx - 10 - 5x)</h3><h3>This will give you 3x^2 + wx - 10 - 5x = -3x^2 - 10. </h3><h3>Now you need to cancel out -3x^2 on both sides. </h3><h3>wx - 10 - 5x = -10</h3><h3>Then cancel out -10 from both sides. </h3><h3>wx - 5x = 0</h3><h3>Now factor out the common term. (x) </h3><h3>w - 5 = 0.</h3><h3>giving you the answer w = 5. </h3><h3 /><h3 /><h3>welcome. *yeets*</h3>
6 0
3 years ago
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0
3 years ago
A company made a profit of 75000 over a period of 6 years on an initial investment of 15000 what is the annual roi
IgorC [24]

Answer:

Find annual profit: $75,000/6 = $12,500


ROI = Annual Profit/ initial investment

ROI = $12,500/$15,000 or 83.3%

hope that help




3 0
2 years ago
Read 2 more answers
Simplify (6ab3c)(-abc2).
sp2606 [1]
Sorry, I know it's a little late.The answer is -6a^2 b^4 c^3
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