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3 years ago

7

Which set of equations is enough information to prove that lines a and b are parallel lines cut by transversal f? m∠4 = 110° and

m∠3 = 70° m∠1 = 110° and m∠2 = 110° m∠1 = 110° and m∠3 = 70° m∠2 = 110° and m∠3 = 110°

1 answer:

seropon [69]3 years ago

3 0

Answer:

its c

Step-by-step explanation:

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Five more than the quotient of a number and 3 equals 2.

MakcuM [25]

Answer:

$\large\boxed{\dfrac{n}{3}+5=2\ or\ n:3+5=2}$

Step-by-step explanation:

$\dfrac{n}{3}+5=2\\\\\dfrac{n}{3}-\text{the quotient of a number}\ n\ \text{ and 3}\\\\...+5-\text{five more than}$

If you want a solution:

$\dfrac{n}{3}+5=2$ <em>subtract 5 from both sides</em>

$\dfrac{n}{3}+5-5=2-5$

$\dfrac{n}{3}=-3$ <em>multiply both sides by 3</em>

$3\!\!\!\!\diagup\cdot\dfrac{n}{3\!\!\!\!\diagup}=(3)(-3)\\\\n=-9$

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3 years ago

When constructing a prependicular bisector why must the compass opening be greater than 1/2 the length of the segment ?

Tomtit [17]

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You have to draw two circles, centered at the endpoints of the segment, and connect their intersection.

If the radii are less than half the length of the circle, the circles will have no interserction.

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3 years ago

How to find the radius of square that tells you length, width, and height.

Anna [14]

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3 years ago

Select all the correct answers.

Nesterboy [21]

Answer:

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3 years ago

The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

$\frac{10000(0.2)^1}{10000(0.2)^0}$

$\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2$

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

8 0

4 years ago