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3 years ago

How to find the width of a rectangle if it's missing A=21 L=7 W= ?

Mathematics

1 answer:

Radda [10]3 years ago

3 0

Answer:

The width is 3

Step-by-step explanation:

the formula is l×w

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Factorise- 6y-15y^2 HELP

Citrus2011 [14]

We will take the common terms in the expressions-

Here in the expression, 3y is a common term.
So,

$6y - 15y^{2}$

$3y(2 - 5y)$

So, the answer is $3y(2 -5y)$

7 0

3 years ago

Read 2 more answers

HELP ASAP , algebra 1 giving BRAINLIEST , no guessing,

katovenus [111]

Answer:

The answer is B that is the answer

5 0

3 years ago

Read 2 more answers

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by h and take the limit as h approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

8 0

2 years ago

Miko filled two tanks with water. The ratio of the amount of water in Tank A to Tank B is 3 : 2. The total amount of water in bo

Anna71 [15]

Step-by-step explanation:

answer is in the above picture

6 0

2 years ago

In his first year, a math teacher earned $32,000. Each successive year, he

Ivan

Answer:

Step-by-step explanation:

Each successive year, he

earned a 5% raise. It means that the salary is increasing in geometric progression. The formula for determining the nth term of a geometric progression is expressed as

Tn = ar^(n - 1)

Where

a represents the first term of the sequence(amount earned in the first year).

r represents the common ratio.

n represents the number of terms(years).

From the information given,

a = $32,000

r = 1 + 5/100 = 1.05

n = 20 years

The amount earned in his 20th year, T20 is

T20 = 32000 × 1.05^(20 - 1)

T20 = 32000 × 1.05^(19)

T20 = $80862.4

To determine the his total

earnings over the 20-year period, we would apply the formula for determining the sum of n terms, Sn of a geometric sequence which is expressed as

Sn = (ar^n - 1)/(r - 1)

Therefore, the sum of the first 20 terms, S20 is

S20 = (32000 × 1.05^(20) - 1)/1.05 - 1

S20 = (32000 × 1.653)/0.05

S20 = $1057920

3 0

3 years ago