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3 years ago

Math be like 0-0????

Mathematics

1 answer:

pashok25 [27]3 years ago

5 0

Answer: A & C

<u>Step-by-step explanation:</u>

HL is Hypotenuse-Leg

A) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH

a leg from ΔABC ≡ a leg from ΔFGH

Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH

B) a leg from ΔABC ≡ a leg from ΔFGH

the other leg from ΔABC ≡ the other leg from ΔFGH

Therefore LL (not HL) Congruency Theorem can be used.

C) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH

at least one leg from ΔABC ≡ at least one leg from ΔFGH

Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH

D) an angle from ΔABC ≡ an angle from ΔFGH

the other angle from ΔABC ≡ the other angle from ΔFGH

AA cannot be used for congruence.

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Subtract 1/8 - 1/4 - 1/2.

Ivan

Answer:

-5/8

Step-by-step explanation:

First find the least common denominator. It is 8.

1/8- 2/8- 4/8

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3 years ago

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If we sub (3x-5y) from (6x +8y)then the result is plz repy

konstantin123 [22]

Answer:

3x + 13y

Explanation:

6x + 8y − (3x − 5y)

Distribute the Negative Sign

6x + 8y + −1(3x − 5y)

6x + 8y + −1(3x) + −1(−5y)

6x + 8y + −3x + 5y

Combine Like Terms

6x + 8y + −3x + 5y

(6x + −3x) + (8y + 5y)

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3 years ago

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Find the prime factorization of 35

AnnyKZ [126]

<h3>Question:</h3><h3>•Find the prime factorization of 35.</h3>

Answer:

•The Prime Factors of 35 are 1, 5, 7, 35 and its Factors in Pairs are (1, 35) and (5, 7).

<h3>Explanation:</h3>

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2 years ago

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A supplier delivers an order for 20 electric toothbrushes to a store. by accident, three of the electric toothbrushes are defect

evablogger [386]

<span>3 of the 20 toothbrushes are defective, so the initial probability of selecting a defective toothbrush on the first try is 3/20. After that, there will be 2 defective toothbrushes in the remaining 19, so the probability of selecting one of those on the 2nd try will be 2/19. To get the probability of those two events happening sequentially, we multiply the two probabilities. Thus (3/20)*(2/19) = 6/380 = 0.0157, or about 1.6%.</span>

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3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

<u>1) Finding $(\overline{x},\overline{y})$ </u>

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

<u>2) Finding the line through $(\overline{x},\overline{y})$ with slope m.</u>

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

<u>3) Find the squared vertical distances between this line and the three points.</u>

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

<u>4) Find the value of m that makes R minimum.</u>

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$