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3 years ago

What is the value of x given the following image ?

Mathematics

2 answers:

USPshnik [31]3 years ago

5 0

What this guy said ok ok

svet-max [94.6K]3 years ago

5 0

Answer:

x=13

Step-by-step explanation:

the angles shown are vertical angles so they are going to have the same value so to find x we setup an equation and the equation would be

-3x+14=-x-12 and the we solve for x

step 1 add 3x to each side

14=2x-12

step 2 add 12 to each side

26=2x

step 3 divide each side by 2

x=13

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A cone-shaped pile of sawdust has a base diameter of 38 feet, and is 14 feet tall. Find the volume of the pile

Romashka-Z-Leto [24]

Answer:

Given: Diameter of cone = 38 feet and height of cone = 14 feet.

Volume of cone V with radius r is one-third the area of the base B times the height h.

i,e $V = \frac{1}{3} B \cdot h$ = $\frac{1}{3} \pi r^2h$ ......[1]

,where B = $\pi r^2$

First find the radius(r);

Using Diameter(D) = 2r

38 =2r

Divide both side by 2 we get;

$\frac{38}{2} =\frac{2r}{2}$

Simplify:

19 = r

or r =19 feet

Now, substitute the value of r = 19 feet and h = 14 feet in [1] [ Use value of $\pi = \frac{22}{7}$ ]

then, we have:

$V = \frac{1}{3} \pi r^2h = \frac{1}{3} \cdot \frac{22}{7} \cdot (19)^2 \cdot (14)$

V = $=\frac{1}{3}\cdot 22 \cdot 19 \cdot 19 \cdot 2 = \frac{22 \cdot 19 \cdot 19 \cdot 2}{3}$

V = $\frac{15884}{3} =5,294.66667$ ≈ 5,294.67 cubic feet.

therefore, the volume of pile is; ≈ 5,294.67 cubic feet.

4 0

3 years ago

Find the arc-length parametrization of the curve that is the intersection of the elliptic cylinder x 2 + y 2/2 = 1 and the plane

storchak [24]

Answer:

$f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)$

0 ≤ Ф ≤ 4π.

Step-by-step explanation:

since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))

the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.

We take

$f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)$

Note that