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2 years ago

• Solve the application problem.

1 answer:

3 0

Not sure what the length of the rectangle is but to find perimeter of a rectangle you can use the following formula

2w+2l= perimeter

The reason we multiply both the length and width by two is because there are two ends of the width and two ends of the length.

The other formula that would also work is as follows

w+w+l+l= perimeter

Hope it helps

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Factor -4 + 8a - 4a^2

tatiyna

Answer: 4(−a+1)(a−1)

Step-by-step explanation:

7 0

1 year ago

Read 2 more answers

Which of the following is NOT a variation of a Pythagorean identity?

Ipatiy [6.2K]

The expression that is not a variation of the Pythagorean identity is the third option.

<h3>What is the Pythagorean identity?</h3>

The Pythagorean identity can be written as:

$sin^2(x) + cos^2(x) = 1$

For example, if we subtract cos^2(x) on both sides we get the second option:

$sin^2(x) = 1 - cos^2(x)$

Which is a variation.

Now, let's divide both sides by cos^2(x).

$sin^2(x)/cos^2(x) + cos^2(x)/cos^2(x) = 1/cos^2(x)\\\\tan^2(x) + 1 = sec^2(x)\\\\tan^2(x) - sec^2(x) = -1$

Notice that the third expression in the options looks like this one, but the one on the right side is positive. The above expression is in did a variation of the Pythagorean identity, then the one written in the options (with the 1 instead of the -1) is incorrect, meaning that it is not a variation of the Pythagorean identity.

Concluding, the correct option is the third one.

If you want to learn more about the Pythagorean identity, you can read:

brainly.com/question/24287773

3 0

2 years ago

Evaluate 2-(-4) + (-g) where y= 7.

Phantasy [73]

Answer:

Step-by-step explanation:

2 - (-4) + (-y)

Substitute y = 7 to the expression:

2 - (-4) + (-7) = 2 + 4 - 7 = 6 - 7 = -1

4 0

1 year ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

juin [17]

Answer:

The Taylor series of f(x) around the point a, can be written as:

$f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....$

Here we have:

f(x) = 4*cos(x)

a = 7*pi

then, let's calculate each part:

f(a) = 4*cos(7*pi) = -4

df/dx = -4*sin(x)

(df/dx)(a) = -4*sin(7*pi) = 0

(d^2f)/(dx^2) = -4*cos(x)

(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4

Here we already can see two things:

the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.

so we only will work with the even powers of the series:

f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....

So we can write it as:

f(x) = ∑fₙ

Such that the n-th term can written as:

$fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}$

6 0

2 years ago

Match the information on the left with the appropriate equation on the right.

solmaris [256]

Answer:

Step-by-step explanation:

1). Equation of a line which has slope 'm' and y-intercept as 'b' is,

y = mx + b

If slope 'm' = 1 and y-intercept 'b' = -3

Equation of the line will be,

y = x - 3

x - y = 3

2). Equation of a line having slope 'm' and passing through a point (x', y') is,

y - y' = m(x - x')

If the slope 'm' = 1 and point is (-1, 2),

The the equation of the line will be,

y - 2 = 1(x + 1)

y = x + 1 + 2

y = x + 3

x - y = -3

3). Equation of a line passing through two points $(x_1, y_1)$ and $(x_2,y_2)$ will be,

$y-y_1=\frac{(y_2-y_1)}{(x_2-x_1)}(x-x_1)$

If this line passes through (-2, 3) and (-3, 4),

$y-3=\frac{(4-3)}{(-3+2)}(x+2)$

y - 3 = -1(x + 2)

y = -x - 2 + 3

y = -x + 1

x + y = 1

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2 years ago