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2 years ago

I need help, the solid below is made from cubes find it volume

Mathematics

2 answers:

Vikki [24]2 years ago

7 0

Answer:

40cm

Step-by-step explanation:

Volume is LxWxH. The length is 2cm, the width is 5cm, and the height is 4cm, you can find these values by counting each block as a cm. Then you multiply all the values, (2x4x5) and you get 40.

Art [367]2 years ago

5 0

Answer:

$[tex]\purple{\rule{45pt}{7pt}}\purple{\rule{45pt}{999999pt}}[tex]$

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What is the area of the trapezoid

irga5000 [103]

The equation for a trapezoid is A = a+b all over 2 times h

So, you would simply substitute.

A=15

B=12

H=8

15+12/2 (8)=

27/2 (8)=

13.5 (8)=

108.

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2 years ago

Read 2 more answers

Find the measure of the exterior Angle1 (I’m a bit confused on how exterior angles work)

sergejj [24]

Answer:

Step-by-step explanation:

I would do 40 + 30 = 70

180-70 = 110

110 is the last angle bc all angles add to 180 in a triangle

Then 180 - 110 = 70 for angle 1 because their supplementary

I don't know if you could just add the 2 inside angles but I would do it this way to be safe until I know for sure.

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2 years ago

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Nesterboy [21]

Answer:

9:50 A.M.!

Step-by-step explanation:

7 HRS +2 HRS = 9 HRS

30 MINS + 20 MINS = 50 MINS

3 0

2 years ago

Tell me how do you find out how many Calendar days are in 7 standard years

larisa86 [58]

2556.6975 days

Your answer would be

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2 years ago

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is

Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = $\frac{dx}{dt}$

ship B is sailing north at 20 km/h = $\frac{dy}{dt}$

here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find $\frac{dz}{dt}$ at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

$z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\$

$z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h$

derivative first equation w . r. to t we get

$2z\frac{dz}{dt} =$ $-2(130-x)\frac{dx}{dt}$ $+2y\frac{dy}{dt}$

$\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]$

$\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }$

$= \frac{1100}{85.44}\\ = 12.87km/h$

8 0

2 years ago