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2 years ago

What is the perimeter of ABC with vertices at A (-11,4) B(-7,8) and C (-4,4)

Mathematics

1 answer:

Alborosie2 years ago

3 0

Answer:

Step-by-step explanation:

Using the distance formula,

$AB=\sqrt{(-11-(-7))^{2}+(4-8)^{2}}=\sqrt{16+16}=4\sqrt{2}\\AC=\sqrt{(-11-(-4))^{2}+(4-4)^{2}}=7\\BC=\sqrt{(-7-(-5))^{2}+(8-4)^{2}}=\sqrt{4+16}=2\sqrt{5}\\\\$

So the perimeter is $4\sqrt{2}+7+2\sqrt{5}$

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Estimate the value of √70 to the nearest integer. Describe your process.

Anastaziya [24]

Answer:

$\large\boxed{\sqrt{70}\approx8}$

Step-by-step explanation:

$\text{We know:}\\\\\sqrt{a}=b\iff a^2=b,\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\5^2=25$

5 0

3 years ago

Kevin is 3 years older than brendon two years ago Kevin was 4 times as old as brendon . How old is Kevin now

Tpy6a [65]

Answer:

Kevin is 6

Step-by-step explanation:

Let k represent Kevin's age now. Then Brendon's age now is (k-3). Two years ago the relationship of their ages was ...

k-2 = 4((k-3) -2)

k -2 = 4k -20 . . . . . eliminate parentheses, collect terms

3k = 18 . . . . . . . . . . add 20-k

k = 6 . . . . . . . . . . . . divide by 3

Kevin is 6 now.

4 0

3 years ago

Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court

Ksju [112]

Answer:

$(a)\dfrac{92}{117}$

$(b)\dfrac{8}{39}$

$(c)\dfrac{25}{117}$

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

$P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}$

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed) $=P(MW)+P(WM)$

$=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}$

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.
Two women are appointed

P(at least one woman is appointed) $=P(MW)+P(WM)+P(WW)$

$P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}$

In Part B, $P(MW)+P(WM)=\frac{8}{39}$

Therefore:

$P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}$

5 0

3 years ago

Simplify |3 − 11| − (15 ÷ 3 + 2) 2

Vikentia [17]

|-8| - (5 + 2)2
8-(7)2
8-14= -6

Follow order of operations

6 0

3 years ago

The radius of the large sphere is double the radius of the small sphere. How many times is the volume of the large sphere than t

kobusy [5.1K]

The volume of a sphere is $\cfrac{4}{3}\ \pi r^3$

So, if the small sphere has radius $r$ and the largest sphere has radius $2r$ , their volumes are

$V_1 = \cfrac{4}{3}\ \pi r^3,\quad V_2 = \cfrac{4}{3}\ \pi (2r)^3 = \cfrac{4}{3}\ \pi 8r^3$

So, their ratio is

$\cfrac{V_2}{V_1} = \cfrac{\frac{4}{3}\ \pi 8r^3}{\frac{4}{3}\ \pi r^3} = 8$

7 0

3 years ago

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