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2 years ago

Hi everyone can someone please help with this problem?

Mathematics

1 answer:

erik [133]2 years ago

8 0

You need to use the equation (secant)(exterior)=(secant)(exterior).
Learned about this today.

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You shouldn't get this question wrong, right?

uranmaximum [27]

Answer:

You have THREE OR 3 APPLES

4 0

2 years ago

Read 2 more answers

Calculate the max shear of cantilever balcony 2m wide 0.2m deep and 1.5m length.

Vikki [24]

Answer:

14.4 kN

Step-by-step explanation:

cantilever balcony 2m wide 0.2m deep and 1.5m length.

= 2 x 0.2 x 1.5

= 0.6 m³

weight of the balcony

= 0.6 x 24

= 14.4 kN.

therefore the max shear 14.4 kN and is acting at support.

8 0

3 years ago

The reciprocal of my number is one fewer than the sum of the reciprocals of the two largest single digit prime numbers. What is

antoniya [11.8K]

The answer would be 2

7 0

2 years ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$

Dropping the units and simplifying, we get the linear ODE

$A'=10+5\sin\dfrac t4-\dfrac A{10}$

$10A'+A=100+50\sin\dfrac t4$

Multiplying both sides by $e^{10t}$ allows us to identify the left side as a derivative of a product:

$10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}$

$\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}$

$e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt$

Integrate and divide both sides by $e^{10t}$ to get

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}$

The tanks starts off with 30 lb of salt, so $A(0)=30$ and we can solve for $C$ to get a particular solution of

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}$

6 0

3 years ago

He unit circle has a radius of 1 unit and is centered at the origin. It is dilated so that it passes through the point (4, 0).

Advocard [28]

We know that

if the dilated circle <span>passes through the point (4, 0)
so
the new radius is the distance from the origin to point (4,0)
d=</span>√[(4-0)²+0²]-----> d=√16------> d=4 units

the new radius is 4 units
original radius is 1 units

[new radius]=[scale]*[original radius]
scale=new radius/original radius-----> scale=4/1-----> 4

the answer is
<span>the scale factor of dilation is 4</span>

5 0

3 years ago