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Sophie [7]
3 years ago
12

Which of the following is a solution to 3tan^3 x= tan x?

Mathematics
1 answer:
Jet001 [13]3 years ago
7 0

Answer:

A. 60°

Step-by-step explanation:

3tan^3 x= tan x

3tan^2x = 1

tan^2 x = 1/3

tan x = 1/\sqrt{3\\}

x = 60°

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What is the answers to the parts 2a-2d
il63 [147K]
2a) if 1 ft³ weighs 150 lb==>the TOTAL VOLUME =5,000/150 =33.334 ft³

2b) 1 ft³ 1,728 in³. So the TOTAL volume in in³ =33.334 x 1728 = 57,600 in³

c) Volume =(1/3)(πR²).H but R = H then V= (1/3)(πR³).plug V (=57600)

57,600 = 1/3 (πR³) ==> R³ = (3 x 57600) / π ==> R = 38 in

d) Area x thickness = Volume ==> Area x 2 in = 57600 in then:
Are =57600/2 & Area =28,800 in²
5 0
3 years ago
Help pls no links I'll give briliantest to who ever gives me the right answer ​
gladu [14]

Answer:

A. 57 7/9

Step-by-step explanation:

4  1/3 x  3  1/3 = 14  4/9

14 4/9 x 4 = 57 7/9

6 0
2 years ago
How many words of five letters can be created that start with the letter T, contain 2 other consonats and end with 2 vowels? Not
Jet001 [13]

If we're only counting 5 vowels (A, E, I, O, U) and 20 consonants (everything else, minus T), then there are

\dbinom52=\dfrac{5!}{2!(5-2)!}=10

ways of picking the vowels, and

\dbinom{20}2=\dfrac{20!}{2!(20-2)!}=190

ways of picking the consonants.

We want the word to start with T, and we'll allow any arrangement of the other 4 letters, so that the total number of words is

4!\dbinom52\dbinom{20}2=\boxed{45,600}

Keep in mind that this means words like TRIES and TIRES are treated as different.

8 0
3 years ago
Find the midpoint of segment AB given. A (-3, 2) and B (3, -2)
Tasya [4]

\text{The formula of the midpoint of segment AB:}\\\\\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right).\\\\\text{We have the points}\ A(-3,\ 2)\ \text{and}\ B(3,\ -2).\ \text{Substitute:}\\\\x=\dfrac{-3+3}{2}=\dfrac{0}{2}=0\\\\y=\dfrac{2+(-2)}{2}=\dfrac{0}{2}=0\\\\Answer:\ \boxed{(0,\ 0)}

8 0
3 years ago
Someone ask me to solve this.
Semenov [28]
X=2h, y=3k

Substitute these values into equations.

y+2x = 4 ------> 3k+2*2h=4 ----->  3k +4h =4

2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1

We have a system of equations now.

 3k +4h =4          ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1

2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0

4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0

(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0

Denominator cannot be = 0
4h(2-2h)≠0

Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4  or h=-1

Check which
4h(2-2h)≠0
1) h= 3/4  , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.

h=3/4, then  3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then  3k = 4-4*(-1) =8 , 3k=8, k=8/3 

So,
if h=3/4, then  k=1/3,
and if h=-1, then  k=8/3 .




6 0
3 years ago
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