All categories

2 years ago

Help me with this question

Mathematics

1 answer:

enot [183]2 years ago

4 0

Answer:

37.2

Step-by-step explanation:

when you turn the small triangle LMN to its right angle to cover the right angle of KLM, you find that they are similar triangles.

therefore the corresponding side lengths are at the same ratio.

LM/KM = MN/LN

LM = 24

MN = 13

we can get LN via Pythagoras of the small triangle

LN² + MN² = LM²

LN² + 13² = 24²

LN² = 24² - 13² = 576 - 169 = 407

LN = sqrt(407) = 20.174241

now back to our main problem

24/KM = 13/sqrt(407)

24×sqrt(407)/13 = KM = 37.2

You might be interested in

To decrease an amount by 19 ,multplu by

Ivahew [28]

Subtraction doesn't mix with mult. or div. Are you sure you've copied this problem down correctly?

"decrease an amount (x) by 19" would be x - 19 (NO multiplication here)

4 0

3 years ago

a A small and big water pump are used to draw water the small water pump draws water at the rate of v litres per minute express

andrew11 [14]

Answer:

mero tauko hola ansqwr tah chaina malaia

6 0

2 years ago

Identify Like terms : 3y2, 2/3x, -3/4y2, 5x, 12x2

wel

3y^2 and -3/4y^2
5x and 2/3x
12x^2

4 0

3 years ago

Lester's Pizza sells pizza by the slice for lunch. Today, there are 80 slices ready for lunch, including 12 slices of pepperoni

Oxana [17]

Whats the questions your asking?

7 0

2 years ago

Find \tan\left(\frac{17\pi}{12}\right)tan( 12 17π )tangent, left parenthesis, start fraction, 17, pi, divided by, 12, end frac

uranmaximum [27]

One way to do this is to notice

$\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4$

Then

$\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}$

We have

$\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

and since $\tan x$ has a period of $\pi$ ,

$\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1$

and so

$\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3$

7 0

3 years ago

Help me with this question ​

Help me with this question