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kifflom [539]
3 years ago
8

Among the eighth-graders in Michael's school, 70% have siblings. The eighth-grade class is 5/6 the size of the seventh-grade cla

ss.
If 112 eighth-graders have siblings, how many students are in the seventh grade?

A. 128

B. 134

C. 160

D. 192
Mathematics
1 answer:
trapecia [35]3 years ago
5 0

Answer:

D. 192

Step-by-step explanation:

This problem involves a few steps.  First, you need to determine how many eighth graders you have given that 112 represent 70% of the total number of eighth graders.  You can solve this using a proportion:

\frac{70}{100}=\frac{112}{x}

Cross-multiply and divide:  70x = 11200 or 70x/70 = 11200/70 or x = 160

Since there are 160 total eighth graders and this is \frac{5}{6} of the amount of seventh graders, we can set up an equation to solve for the total amount of seventh graders:

160 =  \frac{5}{6}x

Multiply by the reciprocal of your fraction on both sides:

\frac{6}{5} x 160 =  \frac{5}{6}x (\frac{6}{5})

Solve for x:  x = 192 seventh graders

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Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1
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\dfrac{\partial f}{\partial x}=-2x=0\implies x=0

\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12

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Now check for extrema on the boundary of D. Convert to polar coordinates:

f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t

Find the critical points of g(t):

\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0

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where n is any integer. There are some redundant critical points, so we'll just consider 0\le t< 2\pi, which gives

t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3

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So altogether, f(x,y) has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

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klasskru [66]

Answers:

  • Similar = Yes, they are similar
  • How:  AA similarity
  • Similarity statement:  triangle HGF ~ triangle HMN

The ~ mark means "similar"

=====================================================

The red angle markings tell us which angles are congruent to one another. We have these two pairs of congruent angles

  • angle H = angle H (both triangles unfortunately reuse H) by the single arc marking
  • angle G = angle M, by the double arc markings

Because we have two pairs of congruent angles, we use the AA similarity theorem. AA stands for angle angle. We could use three pairs, but two pairs is the minimum needed for similarity statements.

-----------

To figure out what goes in the third answer box, note how the order HGF is given to us. The order is important.

We have H first, G second, F third.

H being first, and it having a single arc, means that whatever has a single arc on the other triangle must be first. That would be H. Unfortunately they reuse angle H which makes things confusing a bit.

G is second in HGF, and G has the double arc marking. Note how M also has this double arc marking. Therefore we have M second in the answer. So far we have HM as part of the third answer.

The only thing left is N. Therefore, we have triangle HMN as the answer for the third box.

Put another way, we have these pairings:

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This is why the order is important. So we can see how the angles pair up in the order that they do.

This is why triangle HGF is similar to triangle HMN.

We can write that as triangle HGF ~ triangle HMN

The ~ mark means "similar"

We don't have enough information to prove that the triangles are congruent or not. So we cannot say \triangle HGF \cong \triangle HMN. Your teacher made a typo when using the congruence symbol \cong

Instead, it should be \triangle HGF \sim \triangle HMN

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