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3 years ago

What is the degree of polynomial 3x⁴-4x³y+6x²y²-8x²y³-7y⁴+2

Mathematics

1 answer:

lbvjy [14]3 years ago

7 0

Answer:

Step-by-step explanation:

Find the degree of each monomial term of the polynomial. The highest degree is the answer.

deg(3x^4)=4

deg(-4x^3y)=3+1=4

deg(6x^2y^2)=2+2=4

deg(-8x^2y^3)=2+3=5

deg(-7y^4)=4

deg(2)=0

The highest degree there is 5 so that is the degree of the polynomial.

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How to find the answer to0.291×0.34

Alborosie

Answer:

0.291×0.34=0.09894

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2 years ago

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Multiply and simplify. (2x - 1)(6x - 7)

Kryger [21]

Answer:

D) 12x² - 20x + 7

Step-by-step explanation:

Use FOIL when multiplying 2 binomials...

FOIL is Firsts, Outsides, Insides, Lasts. It's the order in which you multiply the numbers in the binomials...

(2x - 1)(6x - 7)

Firsts: (2x)(6x) = 12x²

Outsides: (2x)(-7) = -14x

Insides: (-1)(6x) = -6x

Lasts: (-1)(-7) = 7

Now add them up...

12x² - 14x - 6x + 7

12x² - 20x + 7

8 0

3 years ago

Find the surface area of the regular pyramid.

spayn [35]

Answer:

Answer = 93.6

Step-by-step explanation:

frist multipy 9 x 13 to get 117 than divided it by 2 = 58.5, than times it by 3 since there is three sides.

(58.5 x 3 is 175.5)

Than for the base multipy 7.8 x 9 = 70.2, but since its a tri, divided it by 2 = 35.1.

Add 58.5 and 35.1 to get 93.6

5 0

2 years ago

Read 2 more answers

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

Find c, the hypotenuse, rounded to four decimal places, in a right triangle with b = 16 and B = 55

Fudgin [204]

Answer:

c=19.5324

Step-by-step explanation:

In given triangle ABC, we have been give values of angle B=55 degree, angle C is right angle and b=16.

Now using those values, we need to find the value of side b.

apply formula of sin

$\sin\left(B\right)=\frac{AC}{AB}$

$\sin\left(55\right)=\frac{16}{c}$

$0.819152=\frac{16}{c}$

$0.819152c=16$

$c=\frac{16}{0.819152}$

$c=19.5323944762$

Hence final answer rounded to four decimal places is c=19.5324

4 0

3 years ago