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3 years ago

Please help this is due in 3 hours!! #17

Mathematics

1 answer:

Westkost [7]3 years ago

4 0

Answer:

multiply 31.4x7.9 then add 9.5 to the answer to multiplication prob. Then add your answer in the first sentence. Then multiply 4.6 x 2 . I think that right,good luck : ) luck.

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Question:

aleksklad [387]

Answer:

There are 10 blocks in a row (section). There are 6 rows (sections).

A. If each block has a value of one, what is the value for a section of blocks?

There are 10 blocks in a section. 10 * 1 = 10

B. What is the total value of all the blocks, write an equation?

There are 6 sections. There are 10 blocks in a section. 6 * 10 * 1 = 60

A. If each block has a value of ten, what is the value for a section of blocks?

There are 10 blocks in a section. 10 * 10 = 100

B. What is the total value of all the blocks, write an equation?

There are 6 sections. There are 10 blocks in a section. 60 * 10 * 10 = 600

4 0

3 years ago

A ship at sea, the Gladstone, spots two other ships, the Norman and the Voyager, and measures the angle between them to be 48°.

OLEGan [10]

Use the sine rule:-

x / sin 48 = 4590 / sin (180-48-55)

x / sin 48 = 4590 / sin 103

x = 4590 * sin 48 / sin 103 = 3500.8 yards to nearest tenth

6 0

3 years ago

Read 2 more answers

Find the missing side lengths leave your answerss as radicals

Phoenix [80]

Answer:

n=7

m= 7 $\sqrt{3}$

Step-by-step explanation:

lmk if you need work

7 0

3 years ago

The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of

Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

$P(t) =504,943.26 -104,943.26e^{0.693t}$

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

$\implies \frac{dP}{dt} =kP$ ----(1)

where k is constant rate at which population is doubled

solving (1)

$ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}$

$t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\$ ---- (2)

initial population = 400,000

population is doubled every week

⇒P(1)=2P(0)

Using (2)

$P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\$

$e^{k} =2\\k=ln|2|\\$

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

$\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\$ ---(3)

solving (3) by calculating integrating factor

$I.F=e^{\int-k dt}$

Multiplying I.F with all terms of (3)

$e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) = -350,000 e^{-kt}$

Integrating w.r.to t

$e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C$

$P(t) =\frac{350,000}{k} +Ce^{kt}\\$

$k=ln|2| =0.693$

$P(t) =504,943.26 + Ce^{0.693t}\\$

at t=0

$P(0) =504,943.26 + Ce^{0.693(0)}$

$400,000 =504,943.26 + C$

$C = -104,943.26$

So, population of mosquitoes in the area at any time t is

$P(t) =504,943.26 -104,943.26e^{0.693t}$

6 0

3 years ago

4. Jason can travel 24 miles in hour . What is his average speed in miles per hour?<br>

Gekata [30.6K]

Answer:

if he is traveling 24 miles in an hour his average speed is going to be 24mph.

3 0

3 years ago

Read 2 more answers