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marusya05 [52]
3 years ago
6

Which equation uses the Pythagorean Theorem to determine whether the side lengths 12, 24, and 26 form a right triangle?

Mathematics
1 answer:
vlada-n [284]3 years ago
3 0

Answer:

C

Step-by-step explanation:

12²+24²=144+576=720≠26²(676)

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If a customer ordered 5 items and the order had a total of 17 wheels, how many wagons were odered?
Nimfa-mama [501]
If a customer ordered 5 items and the order had a total of 17 wheels, we are to determine the number of wagons that was ordered. A wagon has four wheels, therefore, 17 divided by 4 wheels is equal to 4 r. 1. So, 4 wagons were ordered and one extra wheel to complete the 17 wheels and 5 orders.
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Moshe has the container shown below on his desk to hold pencils and paper clips. 2 rectangular prisms. One has a length of 4 inc
iren2701 [21]

Answer:

2 inches by 3 inches by 6 inches and 2 inches by 2 inches by 4 inches

Step-by-step explanation:

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Dan drew the line of best fit on the scatter plot shown below: A graph is shown with scale along x axis from 0 to 10 at incremen
Anit [1.1K]
     0, 3
-   10, 15
= -10, -12
therefore, the slope is 6/5, and the intercept (c) is as supplied, 3.
the equation, y=mx+c or y = a + bx, can be applied here where m or b = 6/5, and a or c = 3.

therefore the equation is y=6/5x+3.

To test this, you can put in y = 10(6/5)+3, which spits out y = 15. This way we know it *should* work.

3 0
3 years ago
You buy 1.86 pounds of ground beef 2,8
Mariana [72]
What is this question asking?
maybe try 3.87
7 0
3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
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