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3 years ago

What is 20 plus 20.

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

4 0

Answer:

Step-by-step explanation:

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

Step-by-step explanation:

+-----

zero plus zero is zero

two plus two is four

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Plugin y=-5 into either equation and solve for x<br> -6x-8y=-20 or x+y=5

nalin [4]

Answer:

x = 10

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

lighting if a 25-foot-tall house cast a 75-foot shadow at the same time that a streetlight casts a 60-foot shadow how tall is th

jolli1 [7]

Answer:

The street light is 20 feet tall.

Step-by-step explanation:

This question depends on a proportion. Keep your height of the object in one ratio and the length of the shadow in the other.

25/x = 75/60 Notice the height of the structure is on the left. The length of the shadow is on the right. Cross multiply

75x = 60 * 25 Combine the right

75x = 1500 Divide by 75

x = 1500/75

x = 20

The street light is 20 feet tall.

3 0

2 years ago

Resuelve el siguiente problema aplicando las estrategias de solución de problemas.

scoundrel [369]

Say it in english so i cant try to help

7 0

2 years ago

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

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

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$

which is about 651.918, so that the area is approximately $651.918\pi\approx\boxed{2048}$ .

Compare this to actual value of the integral, which is closer to 1967.

4 0

3 years ago

Independent Practice

motikmotik

The answer you’re looking for is a yes the domain value five corresponds to two range values -8 and five

5 0

2 years ago