Alec had 750,000. He lost 80%. How many dollars did he lose?

A cube with side length 4p is stacked on another cube with side length 2q^2. What is the total volume of the cubes in factored f

AnnZ [28]

Volume of cube=side³
ok, so you need to know the difference or sum of cubes
a³+b³=(a+b)(x²-xy+y²)
so

(4p)³+(2q²)³=
(4p+2q²)((4p)²-(4p)(2q²)+(2q²)²)=
(4p+2q²)(16p²-8pq²+4q⁴)

3rd option

5 0

2 years ago

Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes

Ivan

Answer:

14.86 knots.

Step-by-step explanation:

Given that:

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2 $\times$ 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 $\times$ 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled $\triangle AOB$ .

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units$

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = $\dfrac{Total\ distance}{Total\ time}$

$\Rightarrow \dfrac{74.28}{5} = 14.86\ knots$

6 0

2 years ago

A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0-min na

Oxana [17]

Answer:

(a) 2.29 km/h

(b) 9 km/h

Step-by-step explanation:

For part (a) you have to apply the average speed formula, which is defined by:

$v=\frac{d}{t}$

where d is the total distance traveled and t is the total time needed.

$d=\frac{4.00}{1.75}=2.29$ km/h

For part (b) you have to calculate the running time (T) , which is the total time of the race minus the nap time:

The nap time in hours is:

90/60 = 1.5 h (because there are 60 minutes in one hour)

The running time is:

T= 1.75 - 1.5 = 0.25 h

Let t1 represent the time before the nap and t2 the time after the nap:

t1+t2 = T

t1+t2 = 0.25

You have to apply the formula d=vt before and after the nap:

-Before the nap, the distance traveled was 0.50 km

0.50 = v1t1

-Afer the nap, the distance traveled was 3.50 km

3.50=v2t2

But v2=2v1 (because after the nap the rabbit runs twice as fast)

You have to solve the system of equations:

t1=0.25-t2 (I)

v1t1=0.50 (II)

2v1t2=3.50 (III)

Replacing (I) in (II)

v1(0.25-t2)=0.50

Applying distributive property and solving:

0.25v1-v1t2=.050

For (III) you have that v1t2=3.50/2=1.75. Hence:

0.25v1-1.75=0.50

Solving for v1:

v1 = 9 km/h

7 0

2 years ago

The price of a laptop is $420. The price of a phone is 35% of the price of this laptop. What is the price of the phone? A $12 B

Pavel [41]

The correct answer would be B. $420x0.35=$147.00

8 0

2 years ago

Read 2 more answers

Jack is flying his kite . He runs 100 feet away from his house and lets out 45 feet of string . The angle of elevation from the

fiasKO [112]

Answer: 125.80 ft

Step-by-step explanation:

Asuming the described situation is as shown in the figure below, we need to find the distance $h$ between the kite and Jacks house, but first we need to find the $x$ , $y$ and then $d$ .

How?

We will use trigonometry, especifically the trigonometric functions sine and cosine:

For $y$ :

$sin(65 \°)=\frac{y}{45 ft}$ (1)

Where $y$ is the opposite side to the angle and $45 ft$ the hypotenuse.

Isolating $y$ :

$y=40.78 ft$ (2)

For $x$ :

$cos(65 \°)=\frac{x}{45 ft}$ (3)

Where $x$ is the adjacent side to the angle.

Isolating $x$ :

$y=19.017 ft$ (4)

Finding $d$ :

$d=x+100 ft$ (5)

$d=19.017 ft +100 ft$

$d=119.017 ft$ (6)

Now that we have found these values, we have to work with a bigger triangle, where the hypotenuse is the distance between the kite and Jack's house $h$ and the sides are the values calculated in (4) and (6).

So, in this case we will use the Pithagorean theorem:

$h^{2}=y^{2} +d^{2}$ (7)

Isolating $h$ and writing with the known values:

$h=\sqrt{y^{2} +d^{2}}$ (8)

$h=\sqrt{(19.017 ft)^{2} +(119.017 ft)^{2}}$ (9)

$h=125.80 ft$ This is the distance between the kite and the house

3 0

2 years ago