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2 years ago

6.

Mathematics

1 answer:

Katarina [22]2 years ago

7 0

Answer:

Step-by-step explanation:

Comparing the triangles in the given questions, the following can be deduced:

5. The scale factor = $\frac{15}{10}$ = $\frac{12}{8}$ ≠ $\frac{9}{7}$

So that;

QR is similar to UT, PR is similar to US, but PQ is NOT similar to TS.

Thus, the triangles are NOT similar.

6. Scale factor = $\frac{15}{10}$ = $\frac{12}{8}$ = $\frac{9}{6}$

= 1.5

So that;

QR is similar to UT, PR is similar to US, but PQ is similar to TS.

Thus, the triangles are similar by Side-Side-Side (SSS).

7. Here, an exact scale factor can not be determined. Thus, the triangles are NOT similar.

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There are 80 pupils in a school. If 32 are boys, what percentage are boys?

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Answer:

40%

Step-by-step explanation:

32/80=.4

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2 years ago

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Two buses leave towns 304 miles apart at the same time and travel toward each other. One bus travels 14 mih slower than the othe

8090 [49]

Answer:

The faster bus moves at 83mi/h and the slower one moves at 69mi/h.

Step-by-step explanation:

Let's define:

R₁ = rate of bus 1, this is the faster one.

R₂ = rate of bus 2, this is the slower one.

We know that one bus travels 14mi/h slower, then:

R₂ = R₁ - 14mi/h.

Now we know that:

Distance = Speed*Time.

If we add the distances that both busses travel in 2 hours, it should be equal to the initial distance between the buses, then:

R₁*2h + R₂*2h = 304 mi

Then we have the two equations:

R₂ = R₁ - 14mi/h

R₁*2h + R₂*2h = 304 mi

The first step is to replace the first equation in the second one, to get:

R₁*2h + (R₁ - 14mi/h)*2h = 304 mi

And now we can solve this for R₁.

R₁*2h + R₁*2h - 14mi/h*2h = 304 mi

R₁*4h - 28mi = 304mi

R₁*4h = 304mi + 28mi = 332mi

R₁ = 332mi/4h = 83mi/h

The faster bus moves at 83mi/h

And we know that the slower one moves at 14mi/h slower than this, then:

R₂ = R₁ - 14mi/h = 83mi/h - 14mi/h = 69 mi/h

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2 years ago

The function f(t) = t2 + 6t − 20 represents a parabola.

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Answer:

Step-by-step explanation:

Part A:

To complete the square, we will do the first few steps all at once, since they are necessary but not complicated to follow all at once. Set the function equal to 0 first, then move the constant over to the other side:

$t^2+6t=20$

The most important rule for completing the square is that the leading coefficient (the number on the t-squared term) is a positive 1. Ours is, but if it wasn't it would need to be factored out.

Next, take half the linear term, square it, and add it to both sides. Our linear term is 6. Half of 6 is 3, and 3 squared is 9, so we add 9 to both sides:

$t^2+6t+9=20+9$

On the right is simple addition, but on the left side, during this process we created a perfect square binomial that will reflect the h coordinate of the vertex. We will state that binomial and do the addition on the right both at the same time:

$(t+3)^2=38$

If you FOIL out the (t + 3)(t + 3) you will get back the polynomial $t^2+6t+9$

Now we can move the constant back over and set the polynomial back to equal f(t):

$(t+3)^2-38=f(t)$

From this we determine the vertex to sit at (-3, -38)

Part B:

Because this is a positive parabola that opens upwards, the function will have a minimum value (if it was negative and opened upside down, it would have a minimum value). In other words, you know the vertex is a minimum because this is a positive parabola that opens upwards.

Part C:

The axis of symmetry is the line that cuts the parabola in half, creating 2 identical, mirror images. Because this parabola opens upwards, a vertical line is required to cut it into 2 identical halves. The vertical line has an equation " x = ". In the case of a positive parabola that opens upwards, the line of symmetry, what x = , will ALWAYS be the number that is inside the parenthesis with the x. Since the standard vertex form is

$f(t)=(t-h)^2+k$