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3 years ago

*WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER**

Mathematics

1 answer:

FromTheMoon [43]3 years ago

6 0

Pythagorean theorem

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Which statement about the ordered pairs (–2, 9), (0, 9) and (3, 9) is not true?

Lelu [443]

Answer:

The slope of these coordinates is 0.

It is a horizontal line.

Step-by-step explanation:

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3 years ago

How many millimeters is in 23 centimeters?

Natalka [10]

(23 centimeters) x (10 millimeters/1 centimeter) = <em>230 millimeters</em>

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3 years ago

Sarah has 25 coins that are dimes and nickels. Together they total $2.00. How many of each type does she have?

Ivanshal [37]

Answer:

15 dimes and 10 nickels

Step-by-step explanation:

15 dimes because 15 multiplied by 10 cents is $1.50

10 nickels because 10 multiplied by 5 cents is $0.50

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3 years ago

Read 2 more answers

Look at the triangle show on the right. The Pythagorean Theorem states that the sum of the squares of the legs of a right triang

Vladimir [108]

Answer:

$cos^2\theta + sin^2\theta = 1$

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

Going by the Pythagoras theorem, we can assume the following.

a = Opposite
b = Adjacent
r = Hypothenuse

So, we have:

$Sin\theta = \frac{a}{r}$

$Cos\theta = \frac{b}{r}$

Having said that:

The expression can be further simplified as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$

Substitute values for sin and cos

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$ becomes

$cos^2\theta + sin^2\theta = 1$

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4 years ago

Subset of {a b c d e f} needs 64 subsets

spin [16.1K]

Answer:

Determine the power set of S, denoted as P:

Term Number Binary Term Subset

12 01100 {b,c}

13 01101 {b,c,e}

14 01110 {b,c,d}

15 01111 {b,c,d,e}

Step-by-step explanation:

please mark my answer in

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3 years ago

***WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER****

*WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER**