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3 years ago

The ratio of the width of the rectangle to the height of the rectangle is ?

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

The area of a rectangle is 108cm (squared. the ratio of the width to the length is 3:4

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Correct Answers only!

Vsevolod [243]

Step-by-step explanation:

Using the given formula :

I = P × R × T

I = 80000 × 15/100 × 4

I = $48000

He saves = $80000 + 48000 = $128000

3 0

3 years ago

Match each value with its formula for ABC.

MariettaO [177]

The solution to the question is:

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

<h3>What is cosine rule?</h3>

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

$c^{2}$ = $a^{2}$ + $b^{2}$ -2ab cos C-----------------1

then c = $\sqrt{a^{2} + b^{2} -2abcosC }$

if b is the side facing the included angle B, then

$b^{2}$ = $a^{2}$ + $c^{2}$ -2accosB-----------------2

b = $\sqrt{a^{2} + c^{2} -2accosB }$

from equation 2, make cosB the subject of equation

2ac cosB = $a^{2}$ + $c^{2}$ - $b^{2}$

cosB = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

if a is the side facing the included angle A, then

$a^{2}$ = $b^{2}$ + $c^{2}$ -2bccosA--------------------3

a = $\sqrt{b^{2} + c^{2} -2bccosA }$

from equation 3, making cosA subject of the equation

2bcosA = $b^{2}$ + $c^{2}$ - $a^{2}$

cosA = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

from equation 1, making cos C the subject

2abcosC = $b^{2}$ + $a^{2}$ - $c^{2}$

cos C = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

In conclusion,

c is 6 = $\sqrt{a^{2} + b^{2} -2abcosC }$

b is 5 = $\sqrt{a^{2} + c^{2} -2accosB }$

cosB is 2 = $\frac{a^{2} + c^{2} - b^{2} }{2ac}$

a is 4 = $\sqrt{b^{2} + c^{2} -2bccosA }$

cosA is 3 = $\frac{b^{2} + c^{2} -a^{2} }{2bc}$

cosC is 1 = $\frac{b^{2} + a^{2} - c^{2} }{2ab}$

Learn more about cosine rule: brainly.com/question/4372174

$SPJ1

4 0

2 years ago

Kayleigh adds two numbers. Each addend has seven digits. The sum has eight digits. What digit is in the ten millions place in th

slava [35]

The answer could be everything from 10-19

6 0

3 years ago

Consider the graph of the parent exponential function fand transformed function g.

zaharov [31]

Answer:

Step-by-step explanation:

4 0

3 years ago

se technology to solve 4x−11=3.2x+13. Enter the solutions in the boxes. Write the lesser solution first. Round to the nearest te

Zielflug [23.3K]

Answer: x=30

Step-by-step explanation: if you were looking for the value of x, hope this helps!

first, you have to make sure that the correct values are on the right side to allow us to find the answer faster and easier. your modified formula should look like this: 4x-3.2x=11+13

if you do the operations on each side, it will look like this: 0.8x=24, from here, all you have to do now is divide 24 by 0.8 to get the X value, which will result in 30!

8 0

3 years ago