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2 years ago

Someone help me please.. I need the slope, y intercept and x intercept

Mathematics

1 answer:

Anuta_ua [19.1K]2 years ago

8 0

Slope is -4x and the y intercept is 20

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Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

5 0

3 years ago

The ratio of pencils to erasers is 4:1. if there are 20 pencils, how many erasers are there?

serious [3.7K]

5 Erasers, 4:1 is equal to 20:5, if you just multiply both numbers by 5

7 0

3 years ago

Read 2 more answers

Graph the following equation: y = 5

Irina-Kira [14]

If you graphed the equation: y = 5

You would put a straight line going horizontally through 5. Like this picture shows. If it were x = 5, for example, then it would go up and down through 5.

Let me know if this helped! Have a nice day!

6 0

3 years ago

Read 2 more answers

First Person = Brainliest.

NeX [460]

Answer:

1080

___ rode a plane for 5^2 miles on a flight to ___. She then took a flight to ___ which was 5^4 miles. How much further did she fly on the second flight?

you can decide where she went and what her name was.

Step-by-step explanation:

6^3 = 213

6^4 = 1296

1296 - 216 = 1080

4 0

2 years ago

Read 2 more answers

A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then s

Bas_tet [7]

Answer:

Step-by-step explanation:

$m=\frac{3}{32}\\ \gamma =0\text{ damping constant.}\\ k=\frac{3}{\frac{1}{4}}=12\\ F(t)=0\\ \text{so equation of motion is }\\ mu''+\gamma u'+ku=F(t)\\ \frac{3u''}{32}+12u=0\\ u''+128u=0\\ \text{and initial conditions are}\\ u(0)=-\frac{1}{12}\\ u'(0)=2\\ u''+128u=0\\ \text{characterstic equation is}\\ r^2+128=0\\ \text{roots are }\\ r=\pm 8\sqrt{2}i\\ \text{therefore general solution is}\\ u(t)=A\cos8\sqrt{2}t+Bsin8\sqrt{2}t\\ \text{using initial conditions we get}\\ A=-\frac{1}{12}\\ B=\frac{\sqrt{2}}{8}\\ \text{therefore solution is }\\ u(t)=-\frac{1}{12}\cos8\sqrt{2}t+\frac{\sqrt{2}}{8}\sin8\sqrt{2}t\\
{hence}\\R=\sqrt{\frac{11}{288}}\\\\\sigma=\pi - \tan^{-1}\frac{1}{\sqrt{2}}\\\\\omega_0=8\sqrt{2}\\T=\frac{4}{4\sqrt{2}}$

3 0

3 years ago