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3 years ago

8

A 5m 60cm vertical pole cast a shadow of 3m 20cm long. Find at the same time the length of a shadow cast by another pole 10m 50c

m high.

1 answer:

Vinvika [58]3 years ago

8 0

Answer:

The answer is 600cm or 6m (I did direct variation)

Plzzz mark me as brainliest

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Im not sure how to solve this

Irina-Kira [14]

A question:
(5.3 x 10 to the 6th power) + 8.2 10 to the 6th power)
A solution:
Multiply 10 x 10 6 times to get rid of the two powers.
(5.3 x 1,000,000) + (8.2 x 1,000,000)
Multiply inside the parenthises.
(5,300,000) + (8,200,000)
Add:
13,500,000.
A answer:
13, 500,000.

B question:
(6.5 x 10 to the 7th power) / (5 x 10 to the 3rd power)
B solution:
Again, get rid of the powers.
(6.5 x 10,000,000) / (5 x 1,000)
Multiply inside the parenthesis:
(65,000,000) / (5,000)
Divide:
13,000
B answer:
13,000.

C question:
(4.6 x 10 to the 6th power) (3 x 10 to the 5th power) / (2 x 10 to the 7th power)
C solution:
Get rid of powers:
(4.6 x 1,000,000) (3 x 100,000) / (2 x 10,000,000)
Multiply:
(4,600,000) (300,000) / (20,000,000)
multiply in the left side:
(1,380,000,000,000) / (20,000,000)
divide:
69,000
C answer:
69,000

8 0

3 years ago

The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

Lemur [1.5K]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Do, integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Substitute the value in 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Hence Cardiac output: $F=0.055 L\s$

4 0

3 years ago

Part 1: Record Information Spend a short amount of time (one to five minutes) doing your activity. Answer the following question

gizmo_the_mogwai [7]

Answer:

I think your gonna have to do this one on your own because this is a whole experiment. Just saying. :)

Step-by-step explanation:

7 0

3 years ago

Maurice puts 130 trading cards in projector sheet he feels 7 sheets and puts the remaining four cards in an 8 sheets have the sa

cestrela7 [59]

Step-by-step explanation:

i am expecting that total cards is 234

7 0

3 years ago

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times sm

sergey [27]

One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle. Then the lines are parallel

<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

$\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}$

And now, 15 degrees is 11 times smaller than the other

Then other angle = 11 times of 15 degrees

$\text {Other angle }=11 \times 15=165 \text { degrees }$

Now, sum of angles = 15 + 165 = 180 degrees.

As we expected their sum is 180 degrees. So the lines are parallel.

Hence, the given lines are parallel

5 0

3 years ago