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2 years ago

The weight of an object on the moon is

Mathematics

1 answer:

Alex Ar [27]2 years ago

7 0

J is a fraction
Rock weighing 202 pounds on earth weighs 201j pounds on the moon.

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HURRY PLEASEEEE

Natasha_Volkova [10]

Answer:

Last one.

Step-by-step explanation:

8 0

2 years ago

What is the dollar amount of the Gentry's annual food and housing budgets combined?

Pavlova-9 [17]

Answer:

The combined amount is $14871.21.

Step-by-step explanation:

It is given that, the total yearly income of gentry family is $29742.42

The % charge on income for food is 20%.

The % charge on income for housing is 30%.

The amount in dollar from % form is given by,

= $(\frac{percent form}{100})(annual income)$

For annual food,

= $(\frac{20}{100})(29742.42)$

= $5948.484

For housing budgets,

= $(\frac{30}{100})(29742.42)$

= $8922.726

Thus the combined amount is,

= $8922.726 + 5948.484 = 14871.21$

= $14871.21

3 0

2 years ago

If we add the second equation to the first equation, we have a new

stiv31 [10]

Answer:

Step-by-step explanation:

7 0

3 years ago

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

The scale of a map is 1 in. = mi. Find each length on the map

WITCHER [35]

The question gives the relationship to be

$1in=21mi$

We will interpolate to solve the question.

First Question: 147 mi

$\begin{gathered} \text{If 1 in = 21 mi} \\ x\text{ in = 147 mi} \end{gathered}$

Cross multiplying,

$undefined$

7 0

1 year ago