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3 years ago

What ratio is equivalent to 4:10 choose all that apply

Mathematics

1 answer:

lara31 [8.8K]3 years ago

7 0

Answer: there are infinite ratios that equal but here is a few examples

8:20 , 2:5 , 12:30 , 16:40 , 20:50 , 24:60 , 28:70 , 32:80 , 36:90

hope this helps mark me brainliest if it did

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Triangle ABC is similar to Triangle DEF. What is the measure of angle DEF? Refer to image, please show work :)

Ne4ueva [31]

Remember, all the interior angles of a triangle equals 180.
So in triangle ABC, it already gives us two out of the three angles.

First, we add 52 and 36 and that gives us 88.

Now we subtract 88 from 180.

180 - 88 = 92

So that means that ∠ABC is 92°

Now we have to look at triangle DEF.
The equation says that triangle DEF is similar to triangle ABC.

Looking at the picture, triangle DEF is smaller and is flipped upside down.

So that means that ∠ABC ≈ ∠DEF

In similar triangles, the measurements of angles does NOT change. Just the side length.

So since ∠ABC = 92° that means that ∠DEF = 92°

4 0

3 years ago

I need to know what is 2x+4

morpeh [17]

Answer:

Step-by-step explanation:

hope this helps

3 0

3 years ago

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Alright I need The answers for number 27 please reply To This question I posted

lys-0071 [83]

Answer:

C. Sqrt(12^2+9^2)

Step-by-step explanation:

Pythagorean theorem is

x^2+y^2=z^2

So solving for z

z=sqrt(x^2+y^2)

z=sqrt(9^2+12^2)

6 0

4 years ago

Find the area of the following rectangle in square feet. 44 ft 2 11 ft 2 14.7 ft 2 132 ft 2

uysha [10]

The picture in the attached figure

we know that
area of rectangle=base*height
in this problem
base is 22 ft
height is 2 yd

convert yd to ft
1 yd is equal to 3 ft
2 yd--------> 2*3-----> 6 ft

area of rectangle=22*6-----> 132 ft²

the answer is
132 ft²

8 0

3 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