All categories

2 years ago

What is the product of 2/12 and 1/14? If you get it correct ill mark ya brainly.

Mathematics

2 answers:

Whitepunk [10]2 years ago

6 0

Answer:

1/84

Step-by-step explanation:

posledela2 years ago

6 0

Answer:

1/84

Step-by-step explanation:

2/12 x 1/14

Can be simplified to:

1/12 x 1/7

12 x 7 = 84

1/84

Please let me know if I'm wrong :)

You might be interested in

How much faster does Student B read than Student A?

Romashka-Z-Leto [24]

Answer:

He reads faster by 8 pages.

3 0

2 years ago

PLEASE HELP URGENT WILL MARK BRAINLIEST

vovikov84 [41]

Answer:b) intersecting

7 0

2 years ago

A) b=2 i used cos

ivann1987 [24]

Answer:

sin 30=B/C

1/2=B/C

2B=C

B=C/2

3 0

2 years ago

(WILL GIVE BRAINLIEST)

weeeeeb [17]

Answer:

m<I=57

m<J=57

m<K=66

Step-by-step explanation:

All the angles in a triangle add up to 180 degrees.

m<I= 3x+18

m<K= 5x+1

m<I is congruent to m<J.

We can plug the information we already know into the equation.

3x+18+3x+18+5x+1=180

11x+37=180

Subtract 37 from both sides.

11x=143

x=13

Now, we can find out what the angles are.

m<I= 3x+18

m<I=3(13)+18

m<I=39+18

m<I=57

We know that m<I=m<J, so they are both equal to 57 degrees.

m<J=57

Now for m<K:

m<K=5x+1

m<K=5(13)+1

m<K=65+1

m<K=66

We know this is correct because 57+57+66=180.

4 0

2 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

2 years ago