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2 years ago

If triangle PQR is similar to SQT, find the value of

Mathematics

2 answers:

ankoles [38]2 years ago

5 0

Answer:

x = 31.4

PQ = 58.8

Step-by-step explanation:

Set up a proportion:

SQ/QP = ST/PR

Substitute in the values they give you:

$\frac{x-9}{x-9+x+5} = \frac{8}{21}$

$\frac{x - 9}{2x - 4} = \frac{8}{21}$

$21x - 189 = 16x - 32\\5x = 157\\x = 31.4$

This means that PQ = 2x - 4 = 2(31.4) - 4 = 62.8-4 = 58.8

exis [7]2 years ago

4 0

Answer:

x = 15, PQ = 26

Step-by-step explanation:

x+13/9 = 21/x-9

x^2+4x−117=168

x^2+4x−285=0

(x−15)(x+19)=0

x=15

it cannot be a negative number so x cannot be -19

PQ= x+13

15+13=26

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arlik [135]

Given:

In a right angle triangle θ is an acute angle and $\tan\theta =\dfrac{3}{5}$ .

To find:

The value of $\cos \theta$ .

Solution:

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

We have,

$\tan\theta =\dfrac{3}{5}$

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

$Hypotenuse=\sqrt{Perpendicular^2+base^2}$

$Hypotenuse=\sqrt{(3x)^2+(5x)^2}$

$Hypotenuse=\sqrt{9x^2+25x^2}$

$Hypotenuse=\sqrt{34x^2}$

$Hypotenuse=x\sqrt{34}$

In a right angle triangle,

$\cos \theta=\dfrac{Base}{Hypotenuse}$

$\cos \theta=\dfrac{5x}{x\sqrt{34}}$

$\cos \theta=\dfrac{5}{\sqrt{34}}$

Therefore, the value of $\cos \theta$ is $\dfrac{5}{\sqrt{34}}$ .

8 0

2 years ago

What is the area of the trapezoid? 8 m 110 m 12 m

VLD [36.1K]

Answer:

the area of the trapezoid would be a = 708; if following that 8 m is the area, 110 m is the base, and 12 m is the height (apologizes if this is the answer you are not looking for)

Step-by-step explanation:

as much as I would like to, I'm really not the best at explaining things

7 0

2 years ago

. A hawk flying at a height of 60 feet spots a rabbit on the ground. If the hawk dives at a speed of 55 feet per second, how

Vladimir [108]

Answer:

The hawk reaches the rabbit at t=4.3 sec

Step-by-step explanation:

Let

h ----> is the height in feet

t ----> is the time in seconds

v ---> the initial velocity in feet pr second

s ----- is the starting height

we have

$h=-16t^{2} +vt+s$

when the hawk reaches the rabbit the value of h is equal to zero

we have

$v=55\ ft/sec$

$s=60\ ft$

substitute

$0=-16t^{2} +55t+60$

Solve the quadratic equation by graphing

The solution is t=4.3 sec

see the attached figure

8 0

3 years ago

What is the answer to this problem-k-6-7k+20=-2

STatiana [176]

Answer:

K=2 that is the answer

6 0

2 years ago

The cone and cylinder above have the same radius, r, and height, h. The volume of the cone is 90 cubic centimeters.

exis [7]

Answer:

$\huge\boxed{\sf Volume\ of\ cylinder = 270\ cm^3}$

Step-by-step explanation:

Volume of cone = 90 cm³

Volume of cylinder = ?

Always remember:

Volume of cylinder = 3 × volume of cone

Volume of cylinder = 3 × 90

Volume of cylinder = 270 cm³

$\rule[225]{225}{2}$

5 0

2 years ago