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3 years ago

What is the perimeter of the larger triangles

Mathematics

1 answer:

kirill [66]3 years ago

5 0

It’s 12 cause the smaller triangle is smaller by times 3 so if u multiply all those and the missing one would be 12 cause 4 times 3 is 12

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What is the counter example of when it rains it pours

Ganezh [65]

A counterexample proves something wrong. To disprove "When it rains, it pours," you could give an example of a time when it rains and does not pour. What if it only rains a little? What if it rains frogs? How are you supposed to "pour" frogs? I dunno. This is sort of an open-ended question. I'd go with "It drizzles, but does not pour."

5 0

3 years ago

Recall the scenario about Eric's weekly wages in the lesson practice section. Eric's boss have been very impressed with his work

Solnce55 [7]

Answer:

$1)\quad f(x)=\bigg\{\begin{array}{ll}12x&0\leq x$

2) D: x = [0, 24]

3) R: y = [0, 384]

4) see graph

<u>Step-by-step explanation:</u>

Eric's regular wage is $12 per hour for all hours less than 9 hours.

The minimum number of hours Eric can work each day is 0.

f(x) = 12x for 0 ≤ x < 9

Eric's overtime wage is $18 per hour for 9 hours and greater.

The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).

f(x) = 18(x - 8) + 12(8)

= 18x - 144 + 96

= 18x - 48 for 9 ≤ x ≤ 24

The daily wage where x represents the number of hours worked can be displayed in function format as follows:

$f(x)=\bigg\{\begin{array}{ll}12x&0\leq x$

2) Domain represents the x-values (number of hours Eric can work).

The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.

D: 0 ≤ x ≤ 24 → D: x = [0, 24]

3) Range represents the y-values (wage Eric will earn).

Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.

The minimum hours he can work in one day is 0.

f(x) = 12x

f(0) = 12(0)

= 0

The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>

f(x) = 18x - 48

f(24) = 18(24) - 48

= 432 - 48

= 384

D: 0 ≤ y ≤ 384 → D: x = [0, 384]

4) see graph.

Notice that there is an open dot at x = 9 for f(x) = 12x

and a closed dot at x = 9 for f(x) = 18x - 48

6 0

3 years ago

Based on the information below, choose the correct answers.

lozanna [386]

Since there are 120 payment and the total interest paid was $ 14,644.95. so the average monthly interest can be calculated by:

Average monthly interest = $ 14,644.95 / 120

Average monthly interest = $ 122.04

% total interest = ($ 14,644.95) / $ 39,644.95 x 100

% total interest =36.94 %

8 0

3 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

Please help me answer number 1

beks73 [17]

Step-by-step explanation:

work is shown and pictured

8 0

3 years ago