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ladessa [460]
3 years ago
12

Andre has 45 sheets of paper and needs to buy more. A store has 5 reams of paper, and each ream contains 500 sheets of paper. Th

e number of sheets of paper Andre has after purchasing x reams is represented by a function.f(x)=500x+45What is the practical domain of the function?
a. all integers from 1 to 5, inclusiveb. all real numbers from 1 to 5, inclusivec. {545, 1045, 1545, 2045, 2545}d. all real numbers
Mathematics
1 answer:
Dmitrij [34]3 years ago
3 0

Answer:

c. {545, 1045, 1545, 2045, 2545}

Step-by-step explanation:

Domain of a function is a set consisting of all the possible outcomes of the variable.

f(x)=500 x+45       where x= 1,2,3,4,5

the possible outcomes for this function are:

f(1)=500(1)+45= 545

f(2)=500(2)+45= 1045

f(3)=500(3)+45= 1545

f(4)=500(4)+45= 2045

f(5)=500(5)+45= 2545

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What is the slope-intercept equation for the line below
rusak2 [61]

Answer:

the answer is A

Step-by-step explanation:

7 0
3 years ago
Hey can you please help me posted picture of question
kogti [31]
Answer:
144

Explanation:
In the closet, we have:
6 pairs of pants
8 shirts
3 pairs of shoes

To know the number of possible combinations that could be made from the available objects, all we have to do is multiply all three numbers together.
Therefore:
possible outcomes = number of pants * number of shirts * number of shoes
possible outcomes = 6 * 8 * 3
possible outcomes = 144

This means that the tree diagram would have 144 leaves to represent all possible combinations.

Hope this helps :)
4 0
3 years ago
Read 2 more answers
What is the product of -3 1/4 × -1 1/2​
Dmitry [639]

For this case we must find the product of the following expression:

-3 \frac {1} {4} * - 1 \frac {1} {2} =

So, we have:

(\frac {4 * (3) +1} {4}) * (\frac {2 * (1) +1} {2}) =\\\frac {12 + 1} {4} * \frac {2 + 1} {2} =\\(- \frac {13} {4}) * (- \frac {3} {2}) =

By law of signs of multiplication is fulfilled:

- * - = +\\\frac {13 * 3} {4 * 2} =\\\frac {39} {8}

ANswer:

\frac {39} {8}

4 0
4 years ago
Read 2 more answers
Identify the standard form of the equation by completing the square.
OLEGan [10]

Answer:

\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1

Step-by-step explanation:

<u>Given equation</u>:

4x^2-9y^2-8x+36y-68=0

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

\implies 4x^2-8x-9y^2+36y=68

Factor out the coefficient of the x² term and the y² term.

\implies 4(x^2-2x)-9(y^2-4y)=68

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2

\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36

Factor the two perfect trinomials on the left side:

\implies 4(x-1)^2-9(y-2)^2=36

Divide both sides by the number of the right side so the right side equals 1:

\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}

Simplify:

\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1

Therefore, this is the standard equation for a horizontal hyperbola with:

  • center = (1, 2)
  • vertices = (-2, 2) and (4, 2)
  • co-vertices = (1, 0) and (1, 4)
  • \textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}
  • \textsf{Foci}: \quad  (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)

4 0
2 years ago
Can someone help me with this problem? It’s Special Right Triangles: Decimal Answer. Round to the nearest tenth. Thank you ! 10
Triss [41]

Answer:

h = 1.4

c = 2.8

Step-by-step explanation:

For each problem, remember the special triangle side ratios then use a proportion. To solve, isolate the variable.

For the triangle with the variable h:

Since two of the angles are 45, this is an isosceles triangle. All isosceles triangles have two equal sides that are not the hypotenuse.

In a right isosceles triangle, the ratio for regular side to hypotenuse is 1 to √2.

\frac{1}{\sqrt2} =\frac{h}{2} \\h = 2\frac{1}{\sqrt2} \\h = \frac{2}{\sqrt2} \\h = \frac{2\sqrt2}{2} \\h = \sqrt{2}

h = √2

h ≈ 1.4

For the triangle with the variable c:

The is an equilateral triangle cut in half because the angles are 30 and 60.

The side ratio of altitude to hypotenuse is √3 to 2.

\frac{\sqrt{3} }{2} =\frac{c}{4} \\\sqrt{3} = \frac{c}{2}\\2\sqrt3 = c

c = 2√2

c ≈ 2.8

5 0
4 years ago
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