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2 years ago

HELP ASAp!!! Will give brainlest. Find all solutions of the equation in the interval

Mathematics

1 answer:

love history [14]2 years ago

4 0

Answer:

7/6π, 1/6π

Step-by-step explanation:

We know that tan is sine/cosine, so cot is cosine/sine.

In this case, we know cotθ = √3

As the number is positive, using the unit circle, we now can say that the two solutions are in quadrants I and III.

Taking a look at the unit circle, we can finalize our answer.

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Rich is tossing a normal quarter. He tosses the quarter 25 times and it lands on heads 5 times. If Rich tosses the quarter again

kotegsom [21]

Answer:

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

Pls help me with this equation

Zanzabum

Answer:

x = 20 degree

Step-by-step explanation:

Hope this helps u !!

4 0

3 years ago

An open box with a square base is to be made from a square piece of cardboard 24 inches on a side by cutting out a square from e

laila [671]

Answer:

x=4 Inch

Step-by-step explanation:

Length of the Square = 24 Inches

If a Square of Length x cm is cut out from each corner

Length of the Box = 24-x-x=(24-2x) Inches

Width of the Box =24-x-x=(24-2x) Inches

Height of the box = x inches

Volume of a Cuboid = Length X Width X Height

V(x)= x(24-2x)(24-2x)

Simplifying

V(x)=4x(12-x)(12-x)

To determine the value of x at which V is largest, we take the derivative of V(x) and solve for the critical points.

V(x)=4x(12-x)(12-x)

$V^{'}(x)=12(x-12)(x-4)$

Set the derivative equal to zero to obtain the critical points

$V^{'}(x)=12(x-12)(x-4)=0\\Since \: 12\neq 0\\(x-12)(x-4)=0\\x-12=0 \:or\: x-4=0\\x=12 \:or\: x=4$

x cannot be equal to 12 as it divides the length of the square cardboard into exactly two equal parts.

When x=4

V(4)=4*4(12-4)(12-4)=16*8*8=1024 Cubic Inches

When x=4 Inch, the volume, V of the open box is largest.

3 0

3 years ago

Simplify 7 − 5(3)2 ?

Anastaziya [24]

The answer is 7 - 7.5

you multiply 5 and 3 and get 15 then divide by 2

8 0

3 years ago

find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).

maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).

To find:

The component equation of the plane.

Solution:

The equation of plane is

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Where, $(x_0,y_0,z_0)$ is the point on the plane and $\left< a,b,c\right>$ is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

$-2(x-(-10))+5(y-7)+1(z-5)=0$

$-2(x+10)+5y-35+z-5=0$

$-2x-20+5y-35+z-5=0$

$-2x+5y+z-60=0$

$-2x+5y+z=60$

Therefore, the equation of the plane is $-2x+5y+z=60$ .

4 0

3 years ago