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3 years ago

PLEASE HELPPP HELP HELP HELP HELP ASAPPP MATH AHHHHH

Mathematics

2 answers:

Rom4ik [11]3 years ago

8 0

Answer:

Step-by-step explanation:

$121 - 55 = 66$

posledela3 years ago

6 0

Answer:

66°

Step-by-step explanation:

Angle of UST = 121 - 55

UST = 66°

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Shaun had a collection of 840 matchbox vehicles. Of these vehicles, 15% are four-wheelers and 30% are convertibles. The rest are

lakkis [162]

Answer:

Step-by-step explanation:

15% of 840 is 126

30% of 840 is 252

252

+ 126

378

840 - 378 = 462

--------------------------------------------------------------------------------------------------------------

55% of 840 is...

462

--------------------------------------------------------------------------------------------------------------

Hope this helps! <3

5 0

3 years ago

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A square with side lengths of 3 cm is reflected vertically over a horizontal line of reflection that is 2 cm below the bottom ed

oksian1 [2.3K]

Answer:

a) 4 cm

b) 5 cm

c) 10 cm

Step-by-step explanation:

The side lengths of the reflected square are equal to the original, and the distance from the axis(2) also remains the same. From there, it is just addition.

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5 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

Pls help or your mom will kidnap you

andrey2020 [161]

I think it is option 1 sorry if wrong

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3 years ago

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