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kumpel
10 days ago
7

Does anyone have the answers to "INVERSES OF LINEAR FUNCTIONS COMMON CORE ALGEBRA II HOMEWORK"

Mathematics
1 answer:
Leona [12.1K]10 days ago
3 0
hopefully this is useful:)) download Socratic, it’s really helpful!!
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Drag and drop an answer to each box to correctly complete the explanation for deriving the formula for the volume of a sphere.
lawyer [12118]
The formula for the volume of a sphere can be derived as follows. We will approach this through calculus, utilizing the concept of a solid of revolution; this is a three-dimensional shape formed by rotating a two-dimensional curve around a straight line (the axis of revolution) that lies within the same plane. From calculus, we know that we will generate a shape by rotating the specified circumference. Next, we isolate y and utilize certain limits for this integral.
5 0
16 days ago
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The difference of two supplementary angles is 88 degrees. Find the measures of the angles.
Zina [11999]
The values of the two supplementary angles are 89 and 1.

To arrive at this, we set the angles as A and B.

We understand that A=B+88 and A+B=90 degrees. Solving this gives A as 89 and B as 1.

8 0
18 days ago
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Consider the vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + u = 8 co
PIT_PIT [11932]

Answer:

u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

Step-by-step explanation:

The characteristic equation is k² + 1 = 0, which leads to k² = -1, resulting in k = ±i.

The roots are k = i or -i.

The general solution takes the form  u(x)=C₁cosx+C₂sinx.

Applying the method of undetermined coefficients, we have

Uc(t) = Pcos wt  + Qsin wt

Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

-Pw^2cos wt  - Qw^2sin wt + Pcos wt  + Qsin wt = 8cos wt.

This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

From -Qw^2 + Q = 8, we can conclude Q = 0.

Thus, Uc(t) = Pcos wt  + Qsin wt = 8 cos wt /(1- w^2).

Combining gives us U(t) = uh(t ) + Uc(t)

     = C1cos t + c2 sin t + 8 cos wt /(1- w^2).

Initial conditions yield:

U(0) = C1cos(0) + c2 sin (0) + 8 cos (0) /(1- w^2)

Which leads us to C1 + 8 /(1- w^2) = 5

So C1 = 5 - 8 /(1- w^2) = -(3 + w^2 ) /(1- w^2).

Next, taking the derivative:

U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2).

Evaluating at t = 0 gives us:

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7.

Thus, c2 = 7.

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

   

4 0
1 month ago
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Which function has a simplified base of 4RootIndex 3 StartRoot 4 EndRoot? f(x) = 2(RootIndex 3 StartRoot 16 EndRoot) Superscript
tester [11911]

Response:

  f(x)=4\sqrt[3]{16}^{2x}

Detailed explanation:

You're likely in search of a function with a base that can be simplified to...

  4\sqrt[3]{4}\approx 6.3496

The functions you seem to be considering appear to be...

  f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x

It looks like the third option is the one that fits your requirements.

9 0
15 days ago
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Len made 1 1/3 liters of punch. He added 2/5 liters of water to make the punch. How much water was used for each liter of punch?
babunello [11329]

Answer:

3/10 liter

Step-by-step explanation:

Assuming your statement indicates that in 4/3 liters of punch, there is 2/5 liters of water, then the proportion of water in the punch is calculated as:

(2/5)/(4/3) = (2/5)(3/4) = 3/10

So, 3/10 of a liter of water is included in each liter of punch.

5 0
1 month ago
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