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2 years ago

What is the inverse of the function f(x) = x + 2?

1 answer:

8 0

Turn f(x) into y so the equation is y=x+2. Now switch their places so it is x=y+2. Now solve for y. The answer is y=x-2.

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Two angles are supplementary. One angle has a measure that is five less than four times the other. What is the measure of the la

iren [92.7K]

Answer:

The larger angle is 143

Step-by-step explanation:

x + y = 180 (This is the meaning of supplementary. Two angles make up 180 degrees)

4x - 5 = y

x + (4x - 5) = 180 Remove the brackets.

x + 4x - 5 = 180

5x - 5 = 180 Add 5 to both sides

5x - 5 +5 = 180 + 5 Combine

5x = 185 Divide by 5

5x/5=185/5

x = 37

y = 4*37 - 5

y = 148 - 5

y = 143

4 0

2 years ago

25. Several functions represent different savings account plans. Which functions are linear?

Mice21 [21]

Answer:

b, c, d

Step-by-step explanation:

y=kx+b

y=kx

so y=5,25x-right

y=0,5x-right

y=2,5x+7-right

8 0

2 years ago

Image 1<br>image 2<br>image 3<br>none

Talja [164]

I’d assume image 1 since it’s a combination of multiple substances or a mixture.

7 0

2 years ago

Hi please help me I'm in the middle of a test and I don't want to get grounded

lawyer [7]

Answer:

8(3j-2)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

8 0

3 years ago