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1 year ago

Which figures can be precisely defined by using only undefined terms? Select three options.

Mathematics

2 answers:

love history [14]1 year ago

8 0

This is a pretty bad question but I think the answer they're looking for is

angle

circle

line segment

We define arcs in terms of circles and parallel lines in terms of lines (though not necessarily line segments, so this is a bit of a judgement call).

Vitek1552 [10]1 year ago

4 0

Angle
Circle
Line segment

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What is the value of the algebraic expression x=1/2, y=1, and z=2? 6x(y2z)

alexandr1967 [171]

Is 12 I think so !! try it

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2 years ago

For ΔABC, ∠A = 4x - 10, ∠B = 5x + 10, and ∠C = 7x + 20. If ΔABC undergoes a dilation by a scale factor of 1 3 to create ΔA'B'C'

VLD [36.1K]

We know that the angles of a triangle sum to 180°. For ΔABC, this means we have:
(4x-10)+(5x+10)+(7x+20)=180

Combining like terms,
16x+20=180

Subtracting 20 from both sides:
16x=160

Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.

For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180

Combining like terms,
20x-20=180

Adding 20 to both sides:
20x=200

Dividing both sides by 20:
x=10

This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.

Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.

5 0

1 year ago

Read 2 more answers

Kim's business earns $10,000 per month. Kim's non-employee expenses are $3,000 per month. If Kim wants $2,000 in profit per mont

MA_775_DIABLO [31]

Answer:

Kim's business earns $10,000 per month.

Kim's non-employee expenses are $3,000 per month.

If Kim wants $2,000 in profit per month, then the maximum amount Kim can spend for employee:

10000 - 3000 - 2000 =5000

If each employee costs $1,000 per month, Kim can recruit 5000/1000 = 5 as the maximum number of employees.

Hope this helps

8 0

1 year ago

Read 2 more answers

A group of campers is going to occupy 3 campsites at a campground. There are 17 campsites from which to choose. In how many ways

Alexeev081 [22]

Answer:

Campsites be chosen in 680 ways.

Step-by-step explanation:

Given:

Number of campsites= 17

Number of campsites that are to be occupied=3

To Find:

Number of ways can the campsites be chosen=?

Solution:

Combination:

In mathematics, a combination is a way of selecting items from a collection where the order of selection does not matter. Suppose we have a set of three numbers P, Q and R. Then in how many ways we can select two numbers from each set, is defined by combination.

nCr = n(n - 1)(n - 2) ... (n - r + 1)/r! = n! / r!(n - r)!

No of ways in which campsites can be chosen= $\frac{17!}{3!(17-3)!}$ 17C3

=> $\frac{17!}{3!(14!)}$

=> $\frac{15\times16\times17}{3\times 2\times 1}$

=> $\frac{4080}{6}$

=>680

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1 year ago

Investigate the following harvesting model both qualitatively and analytically. If a constant number h of fish are harvested fro

zalisa [80]

Answer:

a. The population does not become extinct in finite time.

Step-by-step explanation:

The model for the population of the fishery is

$dP/dt = P(a-bP)-h, P(0) = P_0$

If we rearrange and replace the constants we have:

$\frac{dP}{P(7-P)-49/4} =dt\\\\-4 (\frac{dP}{4(P-7)P+49}) =dt\\\\-4 \frac{dP}{(2P-7)^2} =dt\\\\-4 \int\frac{dP}{(2P-7)^2} =\int dt\\\\-4(-\frac{1}{2(2P-7)})=t+C\\\\\frac{2}{2P-7}=t+C\\\\ t=0 \,\,\, P(0)=P_0\\\\\frac{2}{2P_0-7}=0+C\\\\C=\frac{2}{2P_0-7}$

Now we can calculate if the population become 0 in any finite time

$\frac{2}{2P-7}=t+\frac{2}{2P_0-7}\\\\\frac{2}{2*0-7}=t+\frac{2}{2P_0-7}\\\\-\frac{2}{7}=t+\frac{2}{2P_0-7}\\\\$

To be a finite time, t>0

$t=-\frac{2}{7}-\frac{2}{2P_0-7}=0\\\\-\frac{2}{2P_0-7}=\frac{2}{7}\\\\7-2P_0=7\\\\P_0=0$

We can conclude that the only finite time in which P=0 is when the initial population is 0.

Because P0 is a positive constant, we can say that the population does not become extint in finite time.

5 0

1 year ago