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3 years ago

Help me on this please

Mathematics

2 answers:

Sedbober [7]3 years ago

4 0

The answer would be 125.
The answer is 125 because 10 to the 6th power is 1,000,000 and 10 to the 4th power is 40,000. 5*1,000,000 is 5,000,000. 4*10,000 is 40,000. 5,000,000/40,000 is equal to 125!

serious [3.7K]3 years ago

4 0

Hello,

The answer would be 125.

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Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

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Now multiply both sides by $3x^2$ to get

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so that the left side condenses into the derivative of a product.

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3 years ago

Four pounds of chicken is divided into 1/3 -pound packages. How many packages of chicken are there?

Katen [24]

Answer:12

Step-by-step explanation:

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3 years ago

Tatiana has a thorn in her bike tire but it's not flat yet. She's riding home at a constant speed, and the diameter of her bike

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2 years ago

HELP WOTU FINDING CIRCUMFERENCE PLEASE 20 points!

saw5 [17]

Answer:

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question 2, area=481.609sq meter, lenght of BC=263.3472 meter

for convience i converted yards to meter.

Step-by-step explanation:

in first question to find area of circle by multiplying value of pi (3.14) and square of radius of circle and to find circumference just multiply 2 with value of pi(3.14) and radius of circle.

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8 0

3 years ago

Read 2 more answers

Find (a) the arc length and (b) the area of a sector.

Brut [27]

Answer:

a) 23.56 ft (2 dp)

b) 58.90 ft² (2 dp)

Step-by-step explanation:

<u>Formula</u>

$\textsf{Arc length}=r \theta$

$\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta$

$\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}$

<u>Calculation</u>

Given:

$\theta=\dfrac{3 \pi}{2}$
r = 5 ft

$\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 5\left(\dfrac{3 \pi}{2}\right)\\& = \dfrac{15}{2} \pi \\& = 23.56\: \sf ft\:(2\:dp)\end{aligned}$

$\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(5^2) \left(\dfrac{3 \pi}{2}\right)\\\\& = \dfrac{25}{2}\left(\dfrac{3 \pi}{2}\right)\\\\ & = \dfrac{75}{4} \pi \\\\& = 58.90 \: \sf ft^2\:(2\:dp)\end{aligned}$

6 0

2 years ago