All categories

3 years ago

GEOMETRY HELP PLEASE!! WILL MARK BRAINLEIST! 4

Mathematics

1 answer:

Dmitry [639]3 years ago

5 0

Answer:

5.7 unit

Step-by-step explanation:

From the question,

Applying pythagoras theorem,

a² = b²+c²....................... Equation 1

From the diagram.

a = /QX/, b = /XR/, c = /RQ/

Given: b = /XR/ = 4 units, c = /RQ/ = 4 units.

Substitute these values into equation 1

a² = 4²+4²

a² = 16+16

a² = 32

a = $\sqrt{32}$

a = 5.7 units

Hence, QX = 5.7 units

You might be interested in

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bcos%2030%7D%7B1%2Bsin30%7D%3Dtg30" id="TexFormula1" title="\frac{cos 30}{1+sin30}=tg

Kay [80]

$\cos 30=\frac{\sqrt{3}}{2}$

$\sin 30 = \frac{1}{2}$

$\tan 30=\frac{1}{\sqrt{3}}$

Now we insert that into the equation:

$\dfrac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}} = \frac{1}{\sqrt{3}}$

$\dfrac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\\$

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}\\$

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$

6 0

4 years ago

FREE BRAINLIEST, FIRST ANSWER = BRAINLIEST!

Law Incorporation [45]

Me just so I can ask a question

3 0

3 years ago

Read 2 more answers

A tip employee’s earnings most closely resemble those of which of the following? a. An employee working on straight commission b

Lelechka [254]

Your answer would be C

6 0

3 years ago

Read 2 more answers

Which of these choices show a pair of equivalent expressions?

hoa [83]

Answer:

C AND D

Step-by-step explanation:

AP EX Confirmed

7 0

2 years ago

Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s

Snezhnost [94]

Answer. First option: t > 6.25

Solution:

Height (in feet): h=-16t^2+729

For which interval of time is h less than 104 feet above the ground?

h < 104

Replacing h for -16t^2+729

-16t^2+729 < 104

Solving for h: Subtracting 729 both sides of the inequality:

-16t^2+729-729 < 104-729

-16t^2 < -625

Multiplying the inequality by -1:

(-1)(-16t^2 < -625)

16t^2 > 625

Dividing both sides of the inequality by 16:

16t^2/16 > 625/16

t^2 > 39.0625

Replacing t^2 by [ Absolute value (t) ]^2:

[ Absolute value (t) ]^2 > 39.0625

Square root both sides of the inequality:

sqrt { [ Absolute value (t) ]^2 } > sqrt (39.0625)

Absolute value (t) > 6.25

t < -6.25 or t > 6.25, but t can not be negative, then the solution is:

t > 6.25

5 0

4 years ago

Read 2 more answers