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Irina18 [472]
4 years ago
8

The accompanying line chart shows the major spending categories of the federal budget over the last 50 years.​ (Payments to indi

viduals includes Social Security and​ Medicare; net interest represents interest payments on the national​ debt; all other represents​ non-defense discretionary​ spending.) Complete parts​ (a) through​ (b) below.a. Find the percentage of the budget that went to net interest in 1990 and 2012.The percentage of the budget that went to net interest in 1990 is approximately _______%. Answer – 10 %(Round to the nearest integer as needed.)The percentage of the budget that went to net interest 2012 is approximately _______%. Answer – 5%(Round to the nearest integer as needed.)b. Find the percentage of the budget that went to defense in 1960 and 2012.The percentage of the budget that went to national defense in 1960 is approximately _______%. Answer – 55%(Round to the nearest integer as needed.)The percentage of the budget that went to national defense in 2012 is approximately _______%. Answer – 20%(Round to the nearest integer as needed.)
Mathematics
1 answer:
Galina-37 [17]4 years ago
3 0

Answer:

idk but i think its -5%

Step-by-step explanation:

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Find the coordinates of the other endpoint of the segment, given its midpoint and one endpoint.
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<h2>Answer:</h2><h3>A. Domain </h3>

The domain of a function is the x-values that the graph applies to. This means that the domain is whatever x-values the graph crosses. All vertical parabolas (like the one pictured) have a domain of all reals. This is because any x-value could be plugged into the function and provide a y-value. while it may not seem like it, that graph will cover every single x-value in existence.

  • Domain = All reals

<h3>B. Range</h3>

The range is similar to the domain but is for y-values. So, the range is whatever y-values the graph applies to and crosses. As you can see from the graph, there are no y-values above 1. This means the range is y≤1.

  • Range = y ≤ 1

<h3>C. Increasing Interval</h3>

A graph is increasing when the y-values are increasing. So, on the parent function of a parabola, the graph increases to the right and decreases to the left. However, this graph is inverted and shifted to the left, so the interval will also be flipped and shifted. In this case, the graph increases from -∞ to 2.

  • Increasing Interval = [-∞, 2]

<h3>D. Decreasing Interval</h3>

The decreasing interval is very similar to the increasing interval. This interval applies when the y-values are decreasing as the x-values increase. For a parabola, the increasing and decreasing intervals always meet at the x-value of the vertex, which is 2 on this graph. The y-values decrease during the interval 2 to ∞.

  • Decreasing Interval = [2, ∞]

<h3>E. Opening</h3>

The direction of a parabola is decided by the sign (+ or -) of the leading coefficient. Positive coefficients open up and negative opens down. As you can see from the graph, the sides of the parabola point downwards. This means that the leading coefficient must be negative.

  • Opening = Down

<h3>F. Min and Max</h3>

A parabola will always only have a min or a max, never both. If a graph opens up it has a min because there is one y-value which is the minimum possible y-value. Graphs that open downwards have a maximum because there is one y-value that is the largest possible. So, this graph has a maximum of 1 because that is the largest possible y-value.

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4 0
2 years ago
Find 10 partial sums of the series. (round your answers to five decimal places.) ∞ 15 (−4)n n = 1
nikklg [1K]
Given

\Sigma_{n=1}^\infty15(-4)^n

The first 10 partial sums are as follows:

S_1=\Sigma_{n=1}^{1}15(-4)^n=15(-4)=\bold{-60} \\  \\ S_2=\Sigma_{n=1}^{2}15(-4)^n=\Sigma_{n=1}^{1}15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\  \\ S_3=\Sigma_{n=1}^{3}15(-4)^n=\Sigma_{n=1}^{2}15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\  \\ S_4=\Sigma_{n=1}^{4}15(-4)^n=\Sigma_{n=1}^{3}15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\  \\ S_5=\Sigma_{n=1}^{5}15(-4)^n=\Sigma_{n=1}^{4}15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}

S_6=\Sigma_{n=1}^{6}15(-4)^n=\Sigma_{n=1}^{5}15(-4)^n+15(-4)^6 \\ =-12,300+15(4,096)=-12,300+61,440=\bold{49,140}

The rest of the partial sums can be obtained in similar way.
7 0
3 years ago
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