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3 years ago

Somebody please explain this I have a major exam soon and I really need to know how to solve problems like this

Mathematics

1 answer:

nevsk [136]3 years ago

5 0

Answer:

y = 2x - 3

Step-by-step explanation:

Step 1: Find the slope. You find the slope by using the formula y2-y1/x2-x1. (2, 1) and (4, 5) are our points. Let's find our values.

y2= 5

y1= 1

x2 = 4

x1 = 2

Plug in the numbers into the equation. 5 - 1 is 4. 4 - 2 is 2. 4/2 is 2. There. Our slope is 2.

Step 2: Find the b value. Keep in mind that an equation in point-slope form is y = mx + b, where m = the slope, and b = the y-intercept. For this part, you will plug in one of the points as x and y to find the b value. let's use point (4, 5)

The equation then becomes 5 = 2(4) + b. As we know, 2 * 4 is 8. Subtract 8 on both sides of the equation to cancel out the right side and isolate the b variable. 5 - 8 is -3. Even if you use the other point for the equation (1 = 2(2) +b), you should get the same answer of -3.

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Agroup of friends were working on a student film with a budget of $500, which they

mina [271]

Answer:

89%

Step-by-step explanation:

500-55=445

445/500=0.89

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2 years ago

Read 2 more answers

--Multiplying Polynomials--Use the given formulas to express the volume of each object as a polynomial.

sasho [114]

32.

V = L*w*h

Where:

V= Volume

L= Length = x+5

w= width = x-2

h= height = 6

Replacing with the values given:

V= (x+5) * (x-2) * 6

V =[ (x*x) + (x*-2) + (5*X) +5*-2) ] * 6

V= [ x^2 - 2x + 5x - 10 ] * 6

V= [ x^2 + 3x - 10] *6

V= (x^2*6) + (3x*6)+ ( - 10 * 6)

V= 6x^2 + 18x - 60

6 0

1 year ago

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:

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

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

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

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