Y=3x+2/p make x the subject of the formula

laila [671]

Do 6/30 and you will get .2lbs a day.

3 0

3 years ago

A The number of tropical fish that an aquarium can hold depends on the volume of the fish tank. The interior dimensions of a fis

Delicious77 [7]

Answer:

7

Step-by-step explanation:

We need to first of all find the volume of the fish tank. The most probable shape the ship tank can have is that of a cuboid.

The volume of a cuboid is l * b * h

And this is equals 30cm * 70cm * 90cm = 189,000 cubic centimeters

The number of fish the tank can contain can be obtained by dividing the volume by the volume occupied by a single fish.

This is 189,000/27000 = 7

5 0

3 years ago

Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v

blagie [28]

U = ( -8 , -8)
v = (-1 , 2 )
the magnitude of vector projection of u onto v =
dot product of u and v over the magnitude of v = (u . v )/ ll v ll

ll v ll = √(-1² + 2²) = √5

u . v = ( -8 , -8) . ( -1 , 2) = -8*-1+2*-8 = -8
∴ (u . v )/ ll v ll = -8/√5
∴ the vector projection of u onto v = [(u . v )/ ll v ll] * [v/ ll v ll]

 = [-8/√5] * (-1,2)/√5 = ( 8/5 , -16/5 )

The other orthogonal component = u - ( 8/5 , -16/5 )
= (-8 , -8 ) - ( 8/5 , -16/5 ) = (-48/5 , -24/5 )

So, u as a sum of two orthogonal vectors will be

u = ( 8/5 , -16/5 ) + (-48/5 , -24/5 )

3 0

3 years ago

Read 2 more answers

If two lines are perpendicular, what is the measure of the vertical angles?

Rashid [163]

90 degrees because perpendicular lines are two or more lines that intersect at a 90-degree angle.

5 0

3 years ago

Read 2 more answers

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

3 years ago