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2 years ago

Pls help me with this

Mathematics

2 answers:

yuradex [85]2 years ago

5 0

I think it is 1/18 :)

Akimi4 [234]2 years ago

4 0

I think it’s C. 2 bins

hope this helps! sorry if i’m wrong:)

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Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

son4ous [18]

Answer:

Step-by-step explanation:

Length=18

Width/Breadth =10

Height=2

The formula for a diagonal face of a cuboid is=√(l²+ b²)

Face diagonal=√(18²+10²)

= 20.59 to 2.d.p

The formula for the body diagonal of a cuboid is=√(l² + b² + h²)

Body diagonal=√(18² + 10² + 2²)

=20.69 to 2.d.p

8 0

2 years ago

Please help Meeeeeeeeeee

sweet [91]

Answer:

9 ≤ p

Step-by-step explanation:

-18 ≤ -3p

-18/-3 ≤ p

9 ≤ p

6 0

1 year ago

Read 2 more answers

Find the exact length of the third side

devlian [24]

You can use the Pythagorean theorem, a^2+b^2=c^2, to solve this.

b= square-root of (c^2)-(a^2)

b= square-root of (9^2)-(2^2)

b= 8.77.... which in the exact form is square root of 77.

5 0

2 years ago

WILL GIVE BRAINLIEST!

expeople1 [14]

Answer:

$Average\:Rate\:Change=-59.4$

Step-by-step explanation:

The given function is $f(x)=120(0.1)^x$ .

We want to find the average rate of change of this function from $x=0$ to $x=2$ .

This is given by:

$Average\:Rate\:Change=\frac{f(2)-f(0)}{2-0}$

This implies that;

$Average\:Rate\:Change=\frac{120(0.1)^2-120(0.1)^0}{2-0}$

$Average\:Rate\:Change=\frac{120(0.1)^2-120}{2}$

$Average\:Rate\:Change=\frac{120((0.1)^2-1)}{2}$

$Average\:Rate\:Change=\frac{60(0.01-1)}{1}$

$Average\:Rate\:Change=60(-0.99)$

$Average\:Rate\:Change=-59.4$

5 0

3 years ago

yan [13]

Answer:i dfont kngodw

Step-by-step explanation:

its probably c

3 0

3 years ago