Please answer by today!!!!!

(look at the images) can i get help on these?

Svetlanka [38]

Answer:

see explanation

Step-by-step explanation:

15

JK + KL = JL , substitute values

2x + 1 + 6x = 81, that is

8x + 1 = 81 ( subtract 1 from both sides )

8x = 80 ( divide both sides by 8 )

x = 10

Thus JK = 2x + 2 = 2(10) + 1 = 20 + 1 = 21

KL = 6x = 6 × 10 = 60

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16

JK + KL = JL , substitute values

2x + x + 2 = 5x - 10 , that is

3x + 2 = 5x - 10 ( subtract 3x from both sides )

2 = 2x - 10 ( add 10 to both sides )

12 = 2x ( divide both sides by 2 )

6 = x

Thus

JK = 2X = 2 × 6 = 12

KL = x + 2 = 6 + 2 = 8

JL = 5x - 10 = 5(6) - 10 = 30 - 10 = 20

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17

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = S(7, 3) and (x₂, y₂ ) = T(1, - 5)

d = $\sqrt{(1-7)^2+(-5-3)^2}$

= $\sqrt{(-6)^2+(-8)^2}$

= $\sqrt{36+64}$

= $\sqrt{100}$ = 10

3 0

3 years ago

A city averages 14 hours of daylight in June (the longest days) and 10 in December (the shortest days). Assume that the number o

UNO [17]

The equation is given as

$y = 12 sin (6t - \pi /2) + 2$

<h3>The Amplitude </h3>

The Amplitude can be calculated as

$amplitude = \frac{max + min}{2}\\ amplitude = \frac{10+14}{2} = 12$

<h3>The Vertical Shift</h3>

The vertical shift is the average difference between the maximum and minimum.

$Vertical Shift = \frac{Max - Min}{2} = \frac{14-10}{2} = 2$

For b value

The Time Period will be twice the distance between max and min.

$time period = \frac{2\pi }{b} \\time period = 2 * distance between max and min\\b = \frac{2\pi }{2}*\frac{1}{b} = \frac{\pi }{6}$

<h3>The Period</h3>

The period is calculated as

$T = \frac{2\pi }{b} \\T = \frac{2\pi }{\frac{\pi }{b} } = 12$

The equation is given as

$y = 12 sin (6t - \pi /2) + 2$

Learn more on time period and amplitude here;

brainly.com/question/15531840

8 0

2 years ago

On Monday, a company stock closed at a price of $30.80 per share. On Tuesday, it increased by $1.20 per share. On Wednesday, it

anygoal [31]

If I did this correctly, the answer should be 2.72. Monday's stock price is given to you already, 30.80. To find Tuesday's, you'd add 1.20 to the already existing 30.80, that totals to 32. To find the number for Wednesday you would have to get the percentage and move the decimal so you can multiply it, it should end at .0625. You then get that number and multiply it by 32 because it was Tuesday's amount. From that you should get 2. You then subtract the 2 and the total for Wednesday is 30. You do the same for Thursday, you get the already existing number, 30, and multiply it by .04. That should end in 1.2. You then subtract that 1.2 and you should get 28.8 as your answer for Thursday. For Friday, the steps repeat. You take the 28.8 and multiply it by .025. That should equal to .72. You now subtract that .72 from Thursday's amount and should be left with 28.08 for Friday. Now that all the numbers are known, take the numbers for Monday and for Friday and then subtract them. (30.80-28.08) It should equal to 2.72.

7 0

3 years ago

2 What is the cause of the economie problem facing all countries?

anyanavicka [17]

the answer is uneven distribution of income and wealth

3 0

3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

a)

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

8 0

3 years ago