Which could be the side lengths of a right triangle?

Mathematics

1 answer:

Rudik [331]3 years ago

7 0

Answer:

52,165,173 are the answers for the question

Step-by-step explanation:

please give me brainlest

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Write the equation in slope-intercept form of the line that contains the points (4, -7) and (0, 5).

LenaWriter [7]

The slope intercept form is y=mx+b
The slope formula is m=(y2-y1)/(x2-x1)
So; 5-(-7) / 0-4= -3
Then you use one of the points to find b; I’ll use 0 and 5 (the first number is x and the second number is y)
5=(-3)(0)+b
5=0+b
5=b
Finally you plug in your two values
Y= -3x+5

Hope this helps :)

4 0

4 years ago

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A student rolls a number cube whose six faces are numbered 1 through 6.

zalisa [80]

Answer:

Step-by-step explanation:

The <u>sample space</u> for this experiment is the set of all possible outcomes.

A student rolls a number cube whose six faces are numbered 1 through 6.

Therefore, all possible outcomes are:

rolled number 1;
rolled number 2;
rolled number 3;
rolled number 4;
rolled number 5;
rolled number 6.

Hence, the sample space is {1, 2, 3, 4, 5, 6}

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3 years ago

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Find the measure of the angle with the red dot (NOT X.)<br> Picture Included. Thank you!

Afina-wow [57]

Step-by-step explanation:

I can't see picture very well but as I see first angle is

-4x+5 and second is -13x+39 ( if it isn't, correct me)

sum of this angles is 180° because they are inner angles

(-4x+5) + (-13x+39) = 180

-4x+5 -13x+39=180

-17x+44=180

-17x = 180-44

-17x= 136

x= -8

angle -4x+5 will be:

-4*-8 + 5= 32+5= 37°

3 0

3 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .