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Volgvan
3 years ago
9

URGENT HELP MEEEEE!!!:D

Mathematics
2 answers:
sp2606 [1]3 years ago
4 0

Answer: C: 3rd degree polynomial wit 3 terms.

Step-by-step explanation: The polynomial has 3 terms and the highest degree is 3.

professor190 [17]3 years ago
4 0

C. Third degree polynomial with three terms.

Terms are, by definition, the number of nonzero coefficients for powers of x. In this case, there are 3 nonzero coefficients (3 terms) because -1=-1x^0. For example, 2x^3 has one term, but 0 has 0 terms.

The degree of a polynomial is the highest power involved in the expression. In this case, it's 4x^3, so the degree is 3.

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Which subset of the rational numbers best describes the dimension of a rectangle ?
Alinara [238K]
Either A or D I believe
6 0
3 years ago
Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2
Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

<u>Given quadratic equation</u>:

5x^2-6x-2=0

<u>Define the variables</u>:

  • a = 5
  • b = -6
  • c = -2

<u>Substitute</u> the defined variables into the quadratic formula and <u>solve for x</u>:

\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}

\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}

\implies x=\dfrac{6 \pm \sqrt{76}}{10}

\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}

\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}

\implies x=\dfrac{6 \pm2\sqrt{19}}{10}

\implies x=\dfrac{3 \pm \sqrt{19}}{5}

Therefore, the exact solutions to the given <u>quadratic equation</u> are:

x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}

Learn more about the quadratic formula here:

brainly.com/question/28105589

brainly.com/question/27953354

3 0
1 year ago
Read 2 more answers
Give the coordinates of the image. ​
Keith_Richards [23]
((There has been a malfunction or this users account may have been deleted)) (deleted) (user)
4 0
3 years ago
I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur
Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

to

\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C

Now undo the substitution to get the antiderivative back in terms of x.

=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
4 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%7B10%7D%5Ex%20%3D%205" id="TexFormula1" title=" {10}^x = 5" alt=" {10}^x = 5" align="absmi
My name is Ann [436]

Answer:

below

Step-by-step explanation:

10ˣ = 5

applying log to both sides

xlog 10 = log5

x = 0•699

8 0
3 years ago
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