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3 years ago

CAN SOMEONE CHECK MY WORK GEOMETRY PEOPLE ONLY PLS

Mathematics

1 answer:

quester [9]3 years ago

4 0

Answer:

Is the graph shown the initial triangle, or the translated one?

Step-by-step explanation:

You might be interested in

Can someone please help me Factor 4x^2 - 81

Vsevolod [243]

Answer:

(2x-9)(2x+9)

Step-by-step explanation:

This is a difference of squares because it can be written as (2x)^2-9^2

The formula for factoring a difference of squares is a^2-b^2=(a-b)(a+b)

So replace a with 2x and b with 9 giving us

4x^2-81=(2x-9)(2x+9)

3 0

4 years ago

Read 2 more answers

Roger is a florist who uses a triangular-shaped frame to arrange flower bouquets. He received an order to make a new bouquet suc

kvv77 [185]

The locations specified in the order regarding the shape will be B. longest side AC; smallest angle ∠C.

<h3>How to express the information?</h3>

From the information given, in the triangle where A = 59°, B is unknown, and C = 41°. Therefore, the value of B will be:

= 180° - (59° + 41°)

= 180° - 100°

= 80°

Therefore, the locations specified in the order regarding the shape will be longest side AC; smallest angle ∠C.

In conclusion, the correct option is B.

Learn more about shapes on:

brainly.com/question/25965491

#SPJ1

4 0

2 years ago

Mr.and mrs. Smith's children all play different sports they want to buy a combination of soccer training sets which easr cost $1

shusha [124]

s($149)+v($127)=276
276+f(275)= total
Total $551

3 0

4 years ago

Read 2 more answers

PLEASE HELP!!!!! I WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

Luda [366]

Answer:

Step-by-step explanation:

4 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