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2 years ago

Solve the following equation for a. -4G+1/8Qa=7 Answer: a=?

Mathematics

1 answer:

kolbaska11 [484]2 years ago

6 0

By making a the subject of formula in the given equation, the value "a" is equal to 1/(56Q + 32GQ).

<h3>What is an equation?</h3>

An equation can be defined as a mathematical expression which shows that two (2) or more thing are equal.

In this exercise, you're required to solve for a from the given mathematical expression (equation) by making it the subject of formula. This ultimately implies that, all the other variables would be defined in terms of a and they would all be on the same side of the "equal to" symbol.

Making a the subject of formula, we have:

-4G + 1/8Qa = 7

1/8Qa = 7 + 4G

Multiplying both sides by 8Q, we have:

8Q × 1/8Qa = (7 + 4G) × 8Q

1/a = 56Q + 32GQ

a = 1/(56Q + 32GQ).

Read more on subject of formula here: brainly.com/question/21140562

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Mrs.Jo has $538 class dollars in her classroom bank.Mrs.Hall has $149 less in her classroom bank.How much money does Mrs.Hall ha

Nuetrik [128]

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3 years ago

Read 2 more answers

A population of rabbits is described by the function R(t) = 100(2t/5), where t is measured in months and R is measured in rabbit

Savatey [412]

Answer and Step-by-step explanation: The graph is shown in the attachment.

a. ΔR on [1,2] is mathematically expressed as:

ΔR = R(2) - R(1)

which means difference of population of rabbits after 2 months and after 1 month.

$R(1) = 100(\frac{2}{5}.1 )$

$R(1) = 100(\frac{2}{5} )$

R(2) = $100(\frac{2}{5}.2 )$

$R(2) = 100(\frac{4}{5} )$

$\Delta R = 100(\frac{4}{5} )-100(\frac{2}{5} )$

$\Delta R = 100[\frac{4}{5} - \frac{2}{5} ]$

$\Delta R=$ 40

Difference of rabbits between first and second months is 40.

b. R(0) = 100( $\frac{2}{5} .0$ )

R(0) = 0

Initially, there no rabbits in the population.

c. R(10) = $100(\frac{2}{5}.10 )$

R(10) = 400

In 10 months, there will be 400 rabbits.

d. R(t) = 500

$500=100(\frac{2}{5}.t )$

$\frac{500}{100}=\frac{2}{5}.t$

$t = \frac{500.5}{100.2}$

t = 12.5

In 12 and half months, population of rabbits will be 500.

3 0

3 years ago

A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave

erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$