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3 years ago

Please help thank you and show all work.

Mathematics

1 answer:

Daniel [21]3 years ago

6 0

Question 1:

Look at the first attached image.

We want to solve for the height of the smaller building. Let's start by making a right triangle with angle of 15 degrees and adjacent side of 51 m.

Let's solve for the other leg. Multiply the tangent with 51 m to get the length of the other leg.

$51*tan(15)$
$\approx 14$

Now, subtract that from 207 to get the height of the smaller building.

$207-14=193$

That's your answer.

Question 2:

The angle of elevation is congruent to the angle of depression because they are alternate interior angles.

Thus, you can choose the 1st and 4th choices as your answer.

Question 3:

Look at the second attached image.

Angle E is congruent to angle D for the same reasons mentioned in the last question.
We can create an equation.

$3x+1=2(x+8)$

Distributive property

$3x+1=2x+16$

Subtract both sides by 1 and 2x

$x=15$

Since E and D are congruent, just plug this value into one of the equations for E or D. You get 46.

The third choice is your answer. Hope this helps! :)

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Store A is offering two tubes of lip gloss for $7. The costs for lip gloss at Store B are shown in the table. Assume the cost fo

solmaris [256]

Hi,

I think this is the table, correct?

Number of Tubes 3 5 7

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we can calculate the prize as the number of tubes multiplied by 4:
3*4=12
5*4=20
7*4=28

which means that one tube costs 4 dollars!

However, in the store A two tubes cost 7, which means that one costs half of this -3.5

So Store B is more expensive (4 dollars per 4 rather than 3.5 as in Store A).

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4 years ago

1. What is the area of a circle with a diameter of 3.4 cm? Round your answer to the nearest tenth. (use 3.14 for pi)

Brilliant_brown [7]

1. The area of a circle is
$a = \pi {r}^{2}$
$\pi = 3.14 \\ r = \frac{3.4}{2} = 1.7$
$a = (3.14) {(1.7)}^{2} \\ = 9.0746$
The area is 9.07 cm^2.

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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.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

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3 years ago

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bixtya [17]

Answer:

0.12

hope this is the answer you are looking for

Step-by-step explanation:

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3 years ago

Read 2 more answers

A box contains 8 yellow, 3 orange and 5 purple candies. A tray contains 7 red, 4 blue and 9 orange drinks. If Asia chooses one c

gayaneshka [121]

Okay so we have to first find the total.
8+3+5= 16 candies
7+4+9= 20 drinks
Since she wants to get a yellow candy and any drink but orange, put that information as a fraction.
8/16= 50% or 1/2
11/20= 55% (By the way you get 11 because of the 7 red and 4 blue).
And that's your answer!
Hope this helped and good luck! ^=^

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3 years ago