All categories

3 years ago

A.) 15 B.) 45 C.) 60 D.) 30

Mathematics

1 answer:

Minchanka [31]3 years ago

6 0

The answer is the letter d the anwer is 30 I think it was just a guess and I'm bored dont throw no hate comment cuz I'm from Newark run my 30 just like the answer 30

You might be interested in

What is the excluded value of x / x^2 +2x -3

Rzqust [24]

Answer:

x = - 3 and x = 1

Step-by-step explanation:

Given the rational expression

$\frac{x}{x^2+2x-3}$

The denominator of the expression cannot be zero as this would make the expression undefined. Equating the denominator to zero and solving gives the values that x cannot be, that is

x² + 2x - 3 = 0 ← in standard form

(x + 3)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 1 = 0 ⇒ x = 1

Thus x = 1 and x = - 3 are both excluded values

3 0

3 years ago

SOMEONE HELP ME PLEASE!!!!!!

vaieri [72.5K]

Answer: Choice C

The non-pencil part of the compass is placed at point G. The compass must be kept the same width that was used to form the arc shown. The idea is to draw another arc so that it starts above G (so its on the opposite side of point A) and traces out a similar path as the arc in the diagram.

Check out the attached image to see what I mean. The new second arc is shown in red.

3 0

3 years ago

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

Please help me please

Anastaziya [24]

Answer:

the answer is A ...........

8 0

3 years ago

15 points. Please help meeee

yaroslaw [1]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers