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3 years ago

Given these similar triangles, what is the common ratio of triangle LJK to triangle EFG?

Mathematics

2 answers:

m_a_m_a [10]3 years ago

5 0

Theorem :- In given similar triangle the ratio of corresponding sides are equal !!

answer :- 3 is the common ratio !

kondaur [170]3 years ago

4 0

Find the corresponding sides of each triangle and divide to find the ratio.

FG = KJ

EF = JL

EG = LK

16.8 / 5.6 = 3

20.1 / 6.7 = 3

The common ratio is 3

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!!!PLEASE HELP I'LL GIVE BRAINLIEST AND 30 POINTS!!!

zepelin [54]

If two events are independent, then P(A and B) = P(A) x P(B).

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3 years ago

Question 1 Find the equation for the line that passes through the points (2, 2) and (4, 6). Give your answer in point- slope for

Temka [501]

Answer:

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

Step-by-step explanation:

Equation of a line:

The equation of a line that passes through point $(x_0,y_0)$ , in point-slope form is given by:

$y - y_0 = m(x - x_0)$

In which m is the slope.

Finding the slope:

To find the slope of a line, we need a pair of points. The slope is given by the change in y of these points divided by the change in x.

In this question:

Points (2,2) and (4,6).

Change in y: 6 - 2 = 4

Change in x: 4 - 2 = 2

Slope: $m = \frac{4}{2} = 2$

$y - y_0 = 2(x - x_0)$

I can consider any of these two points $(x_0,y_0)$ , i will use (2,2). So

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

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3 years ago

Problem 5

marysya [2.9K]

Answer: (x, y) = ( -2, -3 )

=============================================

Work Shown:

2x+y = -7

2( x ) + y = -7

2( 4+2y ) + y = -7 .... plug in x = 4+2y

2*4 + 2*2y + y = -7

8 + 4y + y = -7

5y+8 = -7

5y = -7-8 .... subtract 8 from both sides

5y = -15

y = -15/5 ... divide both sides by 5

y = -3

This y value pairs with...

x = 4 + 2y

x = 4 + 2(-3)

x = 4 - 6

x = -2

So we have (x, y) = (-2, -3) as the solution to the system.

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3 years ago

Mean Value theorem help! BRAINLIEST if you can explain it clearly!!!

Korolek [52]

$\bf f(x)=ln(x-5)\qquad [6,8]\qquad \cfrac{df}{dx}=\cfrac{1}{x-5}\\\\
-----------------------------\\\\
\textit{mean value theorem}\qquad f'(c)=\cfrac{f(b)-f(a)}{b-a}\\\\
-----------------------------\\\\
f'(c)=\cfrac{1}{c-5}\qquad thus\quad \cfrac{1}{c-5}=\cfrac{f(8)-f(6)}{8-6}
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\cfrac{1}{c-5}=\cfrac{ln(3)-ln(1)}{2}\implies \cfrac{1}{c-5}=\cfrac{ln(3)-0}{2}
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\cfrac{2}{ln(3)}=c-5\implies \boxed{\cfrac{2}{ln(3)}+5=c}$