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Sonbull [250]
4 years ago
15

What issue results from the combination of limited resources and unlimited wants?

Mathematics
2 answers:
Vinvika [58]4 years ago
8 0

scarcity is the answer your looking for

melomori [17]4 years ago
5 0
<span>The issue that results from the combination of limited resources and unlimited wants? is: Scarcity   </span>
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2. Solve the system using substitution. (1 point)
AveGali [126]

Answer:

(8, 11 )

Step-by-step explanation:

Given the 2 equations

2x + 2y = 38 → (1)

y = x + 3 → (2)

Substitute y = x + 3 into (1)

2x + 2(x + 3) = 38 ← distribute and simplify left side

2x + 2x + 6 = 38

4x + 6 = 38 ( subtract 6 from both sides )

4x = 32 ( divide both sides by 4 )

x = 8

Substitute x = 8 into (2) for corresponding value of y

y = 8 + 3 = 11

Solution is (8, 11 )

7 0
3 years ago
Read 2 more answers
Please help I need help
JulijaS [17]
The rate of change in this pattern is 2:1 comparing the change in 'y' to the change in 'x'.. this means that if you go form 33 to 51 ( an increase of 18), then the the increase from 16 to ? will be 9.  Therefore, ? = 16 + 9 or 25
? = 25
4 0
3 years ago
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
Please help.
34kurt
The number would be greater. if you think about normal multiplication, when you multiply a number by 1 the number stays the same, but when you multiple the number by less then one, i.e. a fraction you get a fraction of what you had. while a number greater then one, i.e. 2, the number increases. so 1.23 is greater then one so it would increase
6 0
3 years ago
Which set of numbers can represent the side lengths, in centimeters, of a right triangle?
ololo11 [35]
The correct answer is B. 10,24,26
7 0
4 years ago
Read 2 more answers
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