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Kryger [21]
3 years ago
9

1. h(2) + g(2) * h(t) = 3-5 g(t) = 2t - 5

Mathematics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

h(2)+g(2) = -3

Step-by-step explanation:

h(2) +g(2)\\ \\h(t) = 3-5, g(t) = 2t -5

        Replace the variable  (t)  with (

2)  in the expression.

h (2) = 3 - 5

        Replace the variable  (t)  with (

2)  in the expression.

g(2) = 2(2) -5

        Replace the function designators in h(2) +g(2) with the actual functions.

h(t) = 3 - 5 +2 (2) ← plug h(2) into 2(t)

         Remove parentheses.

3 - 5 + 2(2)

         Multiply 2 by 2

3 - 5 + 4 - 5

         Subtract  5  from  3

.

-2 + 4 - 5

          Add -2 and 4

2 - 5

          Subtract 5 from 2

-3

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a computer store bought a program at a cost of $20 and sold it at a selling price$26. what is the percent markup
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8 0
3 years ago
How do I graph this?​
vfiekz [6]

See attachment of the graph of the inequalities x + 7y ≤ 49 and 6x + y ≤ 48

<h3>How to graph the inequalities?</h3>

The inequalities are given as:

x + 7y ≤ 49

6x + y ≤ 48

The domain and the range are:

x ≥ 0

y ≥ 0

This means that, we plot the inequalities x + 7y ≤ 49 and 6x + y ≤ 48 under the domain and the range x ≥ 0 and y ≥ 0

See attachment of the graph of the inequalities x + 7y ≤ 49 and 6x + y ≤ 48

Read more about inequalities at:

brainly.com/question/25275758

#SPJ1

6 0
1 year ago
I don't know how to solve it.
Nataly [62]

Step-by-step explanation:

Take the first derivative

\frac{d}{dx} ( {x}^{3}  - 3x)

3 {x}^{2}  - 3

Set the derivative equal to 0.

3 {x}^{2}  - 3 = 0

3 {x}^{2}  = 3

{x}^{2}  = 1

x = 1

or

x =  - 1

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

( - 1) {}^{3}  - 3(  - 1) = 2

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

{1}^{3}   - 3(1)

1 - 3 =  - 2

So the local min occurs at -2, at x=1

8 0
2 years ago
It’s not a question on here but I think you solve for x
Vedmedyk [2.9K]

Answer:

4.56

Step-by-step explanation:

3 0
2 years ago
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