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3 years ago

You have 12 monkey but 5 were taken away how much do you have

Mathematics

2 answers:

sertanlavr [38]3 years ago

8 0

Answer:

12-5=7

unless it's not a prank or a joke question

Gelneren [198K]3 years ago

4 0

Answer:

Step-by-step explanation:

Original number of monkeys = 12

Number taken away = 5

So, number left = 12-5 = 7.

Hope this helps.

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What is the distance between the points (-12, 11) and (-18, 17), to the<br> nearest tenth? *

Basile [38]

Answer:

8.5 units

Step-by-step explanation:

The distance between two points is given by

$d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Taking points (-12, 11) and (-18, 17) as the first and second points respectively then

$d=\sqrt{(-18--12)^{2}+(17-11)^{2}}$

$d=\sqrt {36+36}$

d=8.48528137423857 units

Rounded off to nearest tenths

d=8.5 units

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2 years ago

Find the areas of the shaded regions.

diamong [38]

Answer:

41.8887 cm^2

Step-by-step explanation:

area of a circle
A = pi * r^2
A = pi * (3 + 4)^2
A = pi * 49
A = 153.94

120 / 360 = 1/3
153.94 / 3 = 51.3134

now area of inner circle
A = pi * 3^2
A = pi * 9
A = 28.274
28.274 / 3 = 9.4247

subtract
51.3134 - 9.4247 = 41.8887

7 0

2 years ago

--2, -12, 4, 24, -8, -48, 16, n<br> what's n?

tia_tia [17]

Step-by-step explanation:

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8 0

2 years ago

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Which number is IRRATIONAL? A) 1 B) 125 C) 18 2 3 D) 289 E) 9.4

kodGreya [7K]

An irrational number is one with a fraction.

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2 years ago

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Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

2 years ago