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3 years ago

V - W = uvw, solve for whow do I do that?

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

6 0

Answer:

$W=\frac{V}{UV+1}$

Step-by-step explanation:

$V-W=UVW\\\\V-W+W=UVW+W\\\\V=UVW+W\\\\V=UVW+W\\\\V=W(UV+1)\\\\\frac{1}{UV+1}(V)= \frac{1}{UV+1}(W)(UV+1) \\\\W=\frac{V}{UV+1}$

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If the length of diagonal of a square is 4√2 cm, find it's length, perimeter and area.

Sonbull [250]

Answer:

As Per Provided Information

Length of diagonal of square is 4√2 cm

We have been asked to find the length , perimeter and area of square .

First let's calculate the side of square .

Using Formulae

$\boxed{\bf \:Diagonal_{(Square)} \: = side \sqrt{2}}$

On substituting the value in above formula we obtain

$\qquad\longrightarrow\sf \:4 \sqrt{2} = side \sqrt{2} \\ \\ \\ \qquad\longrightarrow\sf \:4 \cancel{\sqrt{2}} = side \cancel{ \sqrt{2}} \\ \\ \\ \qquad\longrightarrow\sf \:side \: = 4 \: cm$

Therefore,

Length of its side is 4 cm.

Finding the perimeter of square.

$\boxed{\bf \: Perimeter_{(Square)} = 4 \times side}$

Substituting the value we obtain

$\qquad\longrightarrow\sf \:Perimeter_{(Square)} \: = 4 \times 4 \\ \\ \\ \qquad\longrightarrow\sf \:Perimeter_{(Square)} = 16 \: cm$

Therefore,

Perimeter of square is 16 cm .

Finding the area of square .

$\boxed{\bf \: Area_{(Square)} = {side}^{2}}$

Substituting the value we get

$\qquad\longrightarrow\sf \:Area_{(Square)} \: = {4}^{2} \\ \\ \\ \qquad\longrightarrow\sf \:Area_{(Square)} = 16 \: {cm}^{2}$

Therefore,

Area of square is 16 cm².

3 0

2 years ago

Verify cot x sec^4x=cotx +2tanx +tan^3x

Tanzania [10]

Answer:

See explanation

Step-by-step explanation:

We want to verify that:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

Verifying from left, we have

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { \tan}^{2} x )^{2}$

Expand the perfect square in the right:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { 2\tan}^{2} x + { \tan}^{4} x)$

We expand to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + \cot(x){ 2\tan}^{2} x +\cot(x) { \tan}^{4} x$

We simplify to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{2} x}{{ \cos}^{2} x} +\frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{4} x}{{ \cos}^{4} x}$

Cancel common factors:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{{ \sin}x}{{ \cos}x} +\frac{{ \sin}^{3} x}{{ \cos}^{3} x}$

This finally gives:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

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3 years ago

Please help me with this.

tatyana61 [14]

By understanding and applying the characteristics of piecewise functions, the results are listed below:

r (- 3) = 15
r (- 1) = 11
r (1) = - 7
r (5) = 13

<h3>How to evaluate a piecewise function at given values</h3>

In this question we have a piecewise function formed by three expressions associated with three respective intervals. We need to evaluate the expression at a value of the respective interval:

-3 ∈ (- ∞, -1]

r(- 3) = - 2 · (- 3) + 9

r (- 3) = 15

-1 ∈ (- ∞, -1]

r(- 1) = - 2 · (- 1) + 9

r (- 1) = 11

1 ∈ (-1, 5)

r(1) = 2 · 1² - 4 · 1 - 5

r (1) = - 7

5 ∈ [5, + ∞)

r(5) = 4 · 5 - 7

r (5) = 13

By understanding and applying the characteristics of piecewise functions, the results are listed below:

r (- 3) = 15
r (- 1) = 11
r (1) = - 7
r (5) = 13

To learn more on piecewise functions: brainly.com/question/12561612

#SPJ1

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1 year ago

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3 years ago

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2 years ago

V - W = uvw, solve for whow do I do that?​

V - W = uvw, solve for whow do I do that?