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2 years ago

Apply the distributive property to create an equivalent expression, 1/3(3j+ 6)

Mathematics

1 answer:

olchik [2.2K]2 years ago

7 0

Answer:

j + 2

Step-by-step explanation:

1/3(3j) + 1/3(6)

1/3(3j) = j

1/3(6) = 2

Answer: j + 2

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A prop for the theater club’s play is constructed as a cone topped with a half-sphere. What is the volume of the prop? Round you

IrinaVladis [17]

Answer:

1875.6 in³

Step-by-step explanation:

(⅓×3.14×8²×12) + (⅔×3.14×8³)

1875.626667 in³

5 0

3 years ago

Use the Distance Formula to find the distance to the nearest tenth between A(-7, -4) and B(-2, 0).

goldenfox [79]

A(-7,-4) B(-2,0)
√[(x'-x)^2+(y'-y)^2]
√(-2-(-7)^2+(0-(-4)^2
√(5^2)+(4^2)
√25+16
√41

the distance is approximately 6.4 units

8 0

2 years ago

A water-tank has a base with dimensions 2 feet by 6 feet. If a cube with each side 1

Nadya [2.5K]

The increase of volume = volume of the cube = 1 cubic foot = 1728 cu ins
Area of the base = 12 sq ft = 12 * 144 = 1728 sq ins
The height of increase = 1728 / 1728 = 1 inch Answer

7 0

2 years ago

Which function is an example of exponential growth? y = 2(0.3)x y = 1.5(0.4)x y = 3(8)x y = 2(0.7)x

4vir4ik [10]

Exponential equation is
y=a(b)ˣ

when b>1, it is growth
when 0>b>1, it is decay

so

we want the 2nd number to be more than 1
that would be the 3rd one, y=3(8)ˣ

7 0

3 years ago

you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heig

Ostrovityanka [42]

a) $h_0 -u_y t$

b) See interpretation below

Step-by-step explanation:

The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.

Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

$g=32 ft/s^2$

in the downward direction.

For the ball dropped from the initial height of $h_0 = 98 ft$ , the height at time t is given by

$h(t) = h_0 -\frac{1}{2}gt^2$ (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity $u_y$ , and its starting height is zero, so its position at time t is given by

$h'(t)=u_y t - \frac{1}{2}gt^2$ (2)

Therefore, the polynomial that represents the distance between the two balls is:

$h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t$

Now we interpret this polynomial, which is:

$\Delta h(t) = h_0 -u_y t$

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, $h_0$ , is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, $h_0 = 98 ft$

- The coefficient of the linear term, $u_y$ , is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by $u_y$ feet.

5 0

3 years ago