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3 years ago

I have been trying to solve this but couldn’t. please help me!!

Mathematics

1 answer:

Kazeer [188]3 years ago

5 0

I remember being stuck on that question!

the answer is tan x

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The angle measures associated with which set of ordered pairs share the same reference angle

Trava [24]

Answer:

Option 3 is right.

Step-by-step explanation:

Reference angle of x is obtained by either 180-x, 180+x. or 360-x depending on the posiiton of terminal whether II quadrant or iv quadrant, or iii quadrant, etc.

In whatever way we find reference angles,

cos will remain cos only and sin will remain sin only there may be only changes in sign.

Of all the ordered pairs given, we find that I, II, and Iv there is a switch over form cos to sine and sin to cos. Hence these options cannot be for reference angles.

III option is $(-\frac{1}{2} ,\frac{-\sqrt{3} }{2} )(\frac{1}{2} ,\frac{\sqrt{3} }{2} )$

show that both sign and cos changed sign. This is possible only in III quadrant.

ie reference angle of orignal angle t = 180+t

SO this option is right.

7 0

3 years ago

Read 2 more answers

Find the perimeter P of

nataly862011 [7]

Answer:

P = 3.6 + 6 + 3.6 + 6 = 19.2 units

Step-by-step explanation:

distance between (-3,-2) and (-5,-5) =

Sqrt[(-5-3)^2+(-5 - -2)^2]

Sqrt (4+9) = 3.6

d = (-5,-5) and (1,-5) =

Sqrt[(1 - - 5)^2 + (-5 - - 5)^2]

= sqrt (36 + 0) = 6

d = (1,5) and (3,-2)

Sqrt[(1 - - 3)^2 + (-5 - - 2)^2]

Sqrt(4+9) = 3.6

d = (3,-2) and (-3,-2)

Sqrt[(-3-3)^2 + (-2 - - 2)^2]

Sqrt (36+0) = 6

4 0

3 years ago

Are the lines y=3x+10 and y−7=1/3(x+2) parallel, perpendicular, or neither?

olchik [2.2K]

Answer:

Neither

Step-by-step explanation:

The image below shows that the lines aren’t parallel. They instead, intersect, but they are not perpendicular. So it’s neither.

<em>Hope this helps!</em> :)

3 0

3 years ago

Posting for the 3rd time! Someone please help me on this! Thank you

aivan3 [116]

The answer is (-1, 2)

6 0

3 years ago

A closed box with a square base is to have a volume of 171 comma 500 cm cubed. The material for the top and bottom of the box co

Zepler [3.9K]

Answer:

$C(x)=\dfrac{20x^3+1715000}{x}\\$Minimum cost, C(35)=\$29,400$

The dimensions that will lead to minimum cost of the box are a base length of 35 cm and a height of 140 cm.

Step-by-step explanation:

Volume of the Square-Based box=171,500 cubic cm

Let the length of a side of the base=x cm

Volume $=x^2h$

$x^2h=171,500\\h=\dfrac{171500}{x^2}$

The material for the top and bottom of the box costs $10.00 per square centimeter.

Surface Area of the Top and Bottom $=2x^2$

Therefore, Cost of the Top and Bottom $=\$10X2x^2=20x^2$

The material for the sides costs $2.50 per square centimeter.

Surface Area of the Sides=4xh

Cost of the sides=$2.50 X 4xh =10xh

$\text{Substitute h}$=\dfrac{171500}{x^2} $into 10xh\\Cost of the sides=10x(\dfrac{171500}{x^2})=\dfrac{1715000}{x}$

Therefore, total Cost of the box

$= 20x^2+\dfrac{1715000}{x}\\C(x)=\dfrac{20x^3+1715000}{x}$

To find the minimum total cost, we solve for the critical points of C(x). This is obtained by equating its derivative to zero and solving for x.

$C'(x)=\dfrac{40x^3-1715000}{x^2}\\\dfrac{40x^3-1715000}{x^2}=0\\40x^3-1715000=0\\40x^3=1715000\\x^3=1715000\div 40\\x^3=42875\\x=\sqrt[3]{42875}=35$

Recall that:

$h=\dfrac{171500}{x^2}\\Therefore:\\h=\dfrac{171500}{35^2}=140cm$

The dimensions that will lead to minimum costs are base length of 35cm and height of 140cm.

Therefore, the minimum total cost, at x=35cm

$C(35)=\dfrac{20(35)^3+1715000}{35}=\$29,400$

8 0

3 years ago