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3 years ago

The area of a square is given by x2, where x is the length of one side. Mary's original garden was in the shape of a square. She

has decided to double the area of her garden.
Enter an expression that represents the area of Mary's new garden.

Mathematics

1 answer:

Viefleur [7K]3 years ago

3 0

$2( {x}^{2} )$

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Make n the subject of the formula: M=3n

natta225 [31]

Divide each side by 3. ----- n=M/3 .

4 0

3 years ago

Which transformations are needed to change the parent cosine function to y = 3 cosine (10 (x minus pi))?

Alekssandra [29.7K]

Answer:

See Explanation

Step-by-step explanation:

Given

New function: $y = 3 \cos(10 (x -\pi))$

We can assume the parent function to be:

$y = \cos (x)$

The new function can be represented as:

$y = A*\cos((\frac{2\pi}{B})(x-c))$

Where

A = Vertical stretch factor

B = Period

C = Right shift

By comparison:

$y = A*\cos((\frac{2\pi}{B})(x-c))$ to $y = 3 \cos(10 (x -\pi))$

$A = 3$

$c = \pi$

$\frac{2\pi}{B} = 10$

Solve for B

$B = \frac{2\pi}{10}$

$B = \frac{\pi}{5}$

Using the calculated values of $A,\ B\ and\ c.$ This implies that, the following transformations occur on the parent function:

Vertically stretched by $3$
Horizontally compressed by $\frac{\pi}{5}$
Right shifted by $\pi$

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to 3(-5x + 2) + 20x - 7

lukranit [14]

about Answer: The correct answer would be d) 5x - 1

Step-by-step explanation:

Simplify or Evaluate the expression and Combine like terms

3(-5x + 2) + 20x - 7

Learn more about how to solve the problem here: brainly.com/question/17829483

7 0

2 years ago

Read 2 more answers

At the same store, Peter bought 2 pairs of pants and 5 shirts for $61, and Jessica bought 3 pairs of pants and 4 shirts for $67.

svetlana [45]

Hey there!!

Given :

The total cost for 2 pairs of pants and 5 shirts is $61.

The total cost for 3 pairs of paints and 4 shirts is $67.

Let's take the cost for each pair of paints as 'p' and the cost for each shirt as 's'.

Now, let's get these into an equation.

Peter's equation :

... 2p + 5s = 61

Jessica's equation :

...3p + 4s = 67

......................................................................................................

2p + 5s = 61 --- (1)

3p + 4s = 67 --- (2)

Multiply the first equation with 3 and the second equation with 2.

6p + 15s = 183

6p + 8s = 134

Subtract both the equations -

... 7s = 49

Divide both sides with 7.

... s = 7

Hence, the cost of each shirt is $7.

Hope my answer helps!!

6 0

3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

3 years ago