HELLLP- What is the location of the origin? (0,0) (1,1) 0 (10,10)

What is the volume of the regular pyramid below?

mixer [17]

Answer:

A

Step-by-step explanation:

Hopefully this helps

7 0

3 years ago

WORTH 25 POINT PLEASE HELP WILL MARK THE BRAINLIST

AleksandrR [38]

Answer:

sure

Step-by-step explanation:

what is the question

4 0

2 years ago

What is the value of x to the nearest tenth?

Nezavi [6.7K]

Step-by-step explanation:

Hello!!!

Let's workout with this figure.

BC is a chord, O is the centre and OA is the perpendicular bisector.

AB = 1/2 of BC (according to circle's theorem)

so, A B = 1/2 × 25.6

Therefore, the measure of AB is 12.8.

now, let's have a small work with triangle AOB.

as it is a Right angled triangle, taking angle B as refrence angle we get,

p=x

b=12.8

h= OB = 16 (it is also a radius.)

now, 

by Pythagoras relation we get, 

 $p = \sqrt{ {h}^{2} - {b}^{2} }$ 

or, x = root 16^2- 12.8 ^2

by simplification, we get;

the measure of x is 9.6.

Therefore, the value of x is 9.6.

Hope it helps...

8 0

3 years ago

Naty used 91.36 gallons of water over 2.3 days what is the average amount of water naty used?

MAVERICK [17]

39.7 is the average she used

3 0

2 years ago

"Lashonda swam 4 kilometers against the current in the same amount of time it took her to swim 16 kilometers with the current. T

Naddika [18.5K]

Answer:

5 km/h

Step-by-step explanation:

In this problem, Lashonda swam 4 km against the current. So the distance covered in this case is

$d_1=4 km$

Calling $v$ the velocity of Lahonda without the current, and $c$ the velocity of the current, in this situation Lahonda's velocity is

$v-c$

So we can write:

$t_1=\frac{d_1}{v-c}$

where $t_1$ is the time taken to cover the distance.

When Lashonda swims with the current, her velocity is

$v+c$

So we can write

$t_2=\frac{d_2}{v+c}$

where

$d_2=16 km$ is the distance covered in this case, and $t_2$ the time taken.

The velocity of the current is

$c=3 km/h$

Since Lashonad takes the same time to cover the two distances,

$t_1=t_2$

So we can write

$\frac{d_1}{v-c}=\frac{d_2}{v+c}$

And solving for v, we find Lashonda's velocity without the current:

$d_1(v+c)=d_2(v-c)\\d_1 v+d_1c = d_2v-d_2c\\v(d_2-d_1)=c(d_1+d_2)\\v=\frac{d_2+d_2}{d_2-d_1}c=\frac{16+4}{16-4}(3)=5 km/h$

7 0

3 years ago