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3 years ago

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Mathematics

2 answers:

IrinaK [193]3 years ago

8 0

Answer:

4.8

Step-by-step explanation:

JK is twice as long as LM, so is 4.8 units long.

"Similar" means the sides of one figure have the same ratio as the corresponding sides of the other figure. We see that AB = 2×CD. JK corresponds to AB, and LM corresponds to CD.

We could bother to calculate that sides of JKLM are 1.2 times as long as the corresponding sides of ABCD, but that is not necessary. We only need to find a convenient ratio between the side we want and some other known side. I like a ratio of 2, because it is easy to double a number. We find AB=2CD, so JK=2LM=2×2.4 = 4.8.

Vesna [10]3 years ago

6 0

Answer:

4.8

Step-by-step explanation:

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Simplify 12(6)-7(2)+(6-3)

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Step-by-step explanation:

12(6) - 7(2) + (6-3)

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Which choice is the equation of a line that passes through (9, -3) and is parallel to the line represented by this equation? y=5

miskamm [114]

Answer:

$y+3=\frac{5}{3}(x-9)$ or $y=\frac{5}{3}x-18$

Step-by-step explanation:

step 1

Find the slope of the line parallel to the given line

we know that

If two lines are parallel, then their slopes are the same

we have

$y=\frac{5}{3}x-4$

therefore

The slope m of the line parallel to the given line is

$m=\frac{5}{3}$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{5}{3}$

$point\ (9,-3)$

substitute

$y+3=\frac{5}{3}(x-9)$ ----> equation of the line in point slope form

step 3

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$y+3=\frac{5}{3}(x-9)$

Isolate the variable y

$y+3=\frac{5}{3}x-15$

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3 years ago

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Answer:

27 weeks.

Step-by-step explanation:

You need to calculate the minimun common multiple between the three times.

1 week = 1^1 week

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3 years ago

What are the vertical and horizontal asymptotes for the function f(x)=<br> 3x2/x2-4

Alecsey [184]

Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: $f(x)=\dfrac{3x^2}{x^2-4}$

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. $x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2$

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at $y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}$

i.e. $y=\dfrac{3}{1}=3$

Hence, f(x) will have horizontal asymptote at y=3.

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