Can someone help, please?

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

You have to first mess around with the first shape, ABCD, and split that into a rectangle and a right triangle. once you do that, it's pretty painstaking, but simple.

if you look at it you can tell that EFGH is just half the size, but the same ratios and everything.

So, you would just take every perimeter measurement from ABCD, and divide it by two and then sum them together.

2.5 + 1.5 + 4.0 + 2.0 = 10

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A child is hopping along a sidewalk. The ratio table below shows the comparison between the number of hops and the distance trav

natima [27]

First we rewrite the table:
20 30
50 75
80 120
150 ?
We observe that for each 30 number of hops, the distance increases by 45 feet.
We must find the slope of the line:
m = (y2-y1) / (x2-x1)
m = (75-30) / (50-20)
m = 1.5
Then, the line is:
y = 1.5x
We substitute x = 150
y = 1.5 * (150)
y = 225
Answer:
Find the ratio of hops to distance traveled (1: 1.5), then multiply 150 by 1.5.

3 0

3 years ago

Read 2 more answers

For what value of c is the function defined below continuous on (-\infty,\infty)?

kozerog [31]

$f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
$

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
 $\lim_{x \rightarrow 4} \ f(x) = f(4)$

In notation we write respectively
 $\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)$

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
$\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2$

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²

$c^2+4c+4=0
\\(c+2)^2=0
\\c=-2$

That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers $(-\infty, +\infty)$