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2 years ago

Solve for f(-7) plz thanks

Mathematics

1 answer:

Natasha2012 [34]2 years ago

7 0

Answer:

Step-by-step explanation:

If f(x) = 5 - x

Then f(-7) = 5 - (-7)

f(-7) = 5 + 7

f(-7) = 12

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What is 8 × 1 + 3 × 0.1 + 9 × 0.01 in standard form

emmasim [6.3K]

Answer:

0.216

Step-by-step explanation:

$8 \times 1 \times 3 \times 0.1 \times 9 \times 0.01\\\\=(8\times1)(3\times0.1)(9\times0.01)\\\\=8(0.3)(0.09)\\\\=(8\times0.3)(0.09)\\\\=(2.4)(0.09)\\\\=2.4\times0.09\\\\=0.216$

is this the standard form?

5 0

3 years ago

A) x³y²if x = 2 and y = 4

kow [346]

Answer:

128

Step-by-step explanation:

x³y²if x = 2 and y = 4

x³*y²

$(2)^{3} * (4)^{2}$

(2*2*2)*(4*4)

(8)*(16)

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8 0

1 year ago

Please help with the answer below

kenny6666 [7]

Answer:

$f(x) = \frac{x + 7}{6 - x} \\ \frac{3}{2} = \frac{x + 7}{6 - x} \\ 3(6 - x) = 2(x + 7) \\ 18 - 3x = 2x + 14 \\ - 3x - 2x = 14 - 18 \\ - 5x = - 4 \\ \frac{ - 5x}{ - 5} = \frac{ - 4}{ - 5} \\ x = \frac{4}{5}$

3 0

2 years ago

If 2i is a zero of f(x)=x^4+x^2+a, find the value of a

Sedbober [7]

By the polynomial remainder theorem, because x = 2i is a zero of f(x), we have no remainder upon dividing f(x) by x - 2i.

Computing the quotient yields

$\dfrac{x^4+x^2+a}{x-2i} = x^3+2ix^2-3x-6i+\dfrac{12+a}{x-2i}$

Then if the remainder term is 0, follows that a = -12.

5 0

2 years ago

Calculus 2 master needed; stuck on evaluating the integral please show steps <img src="https://tex.z-dn.net/?f=%5Cint%20%7Bsec%2

kakasveta [241]

Answer:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

Step-by-step explanation:

So we have the integral:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx$

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

$y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx$

Therefore, the integral is:

$=2\int \sec(y)\tan^5(y)dy$

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

$=2\int \sec(y)\tan^4(y)\tan(y)dy$

Now, we can use a variation of the trigonometric identity:

$\tan^2(y)+1=\sec^2(y)$

So:

$\tan^2(y)=\sec^2(y)-1$

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

$=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy$

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

$u=\sec(y)\\du=\sec(y)\tan(y)dy$

Substitute:

$2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du$

Expand the binomial:

$=2\int u^4-2u^2+1du$

Spilt the integral:

$=2(\int u^4 du+\int-2u^2du+\int +1du)$

Factor out the constant multiple:

$=2(\int u^4du-2\int u^2du+\int(1)du)$

Reverse Power Rule:

$=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))$

Simplify:

$=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)$

Distribute the 2:

$=\frac{2u^5}{5}-\frac{4u^3}{3}+2u$

Substitute back secant for u:

$=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)$

And substitute back 1/2x for y. Therefore:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)$

And, finally, C:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

And we're done!

7 0

3 years ago

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