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3 years ago

Find the length of side x to the nearest tenth. 30° 12 x х 60°

Mathematics

1 answer:

k0ka [10]3 years ago

8 0

Answer:

13.9

Step-by-step explanation:

Use the Pythagorean Theorem

x=13.85 round to the nearest tenth 13.9

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JK and LM are straight lines. Find /NOK. ( help me pls)

VashaNatasha [74]

Answer:

$NOK = 58$

Step-by-step explanation:

Given

$JOM = 108$

$LON = 50$

Required

Find NOK

From the attached triangle, we have:

$LOK = JOM =108$ --- Corresponding angles

And

$LOK = LON + NOK$

Substitute for LOK and LON

$108= 50+ NOK$

Make NOK the subject

$NOK = 108 - 50$

$NOK = 58$

8 0

3 years ago

1. Solve the quadratic equation using quadratic formula : x2 – 7x + 12 = 0

frez [133]

Answer:

x = 4 and 3

Step-by-step explanation:

[7±√(-7)²-4(1)(12)]/2

x = 4, 3

3 0

3 years ago

HELP MATH GRAPHS AND RULE SOMETHING LIKE THAT

Bess [88]

I dont know math well but try to look it up

4 0

2 years ago

What are the roots of a problem (Math)

puteri [66]

The roots, zeros, or solutions (they all means same thing) of an equation are when you graph that equation, x-intercepts are the real roots, zeros, or solutions.

Or set the y equal to zero which does the same job.

6 0

3 years ago

77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

Colt1911 [192]

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$ .

6 0

2 years ago