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2 years ago

20 points repost because no one answered. Geomertry question

Mathematics

2 answers:

aniked [119]2 years ago

7 0

Answer:

x=2.25

Step-by-step explanation:

ΔABC and ΔADE are similar.

∴ sides are proportional.

$\frac{5}{1} =\frac{x+9}{x} \\5=\frac{x}{x} +\frac{9}{x} \\5=1+\frac{9}{x}\\\frac{9}{x}=5-1\\\frac{9}{x}=4\\x=\frac{9}{4}=2.25$

Ira Lisetskai [31]2 years ago

4 0

Answer:

x = 9/4

Step-by-step explanation:

We know that the triangles are similar

We can use ratios to solve

x x+9

----- = ----------------

1 5

Using cross products

5x = 1(x+9)

5x = x+9

Subtract x

5x-x = x+9-x

4x= 9

Divide by 4

4x/4 = 9/4

x = 9/4

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For f(x) = 2x – 3 find f(-1/4)

Firdavs [7]

9514 1404 393

Answer:

-3 1/2

Step-by-step explanation:

Put the argument where the variable is and do the arithmetic.

f(-1/4) = 2(-1/4) -3 = -1/2 -3

f(-1/4) = -(3 1/2)

5 0

2 years ago

47 + 15 .....................

Igoryamba

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

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

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