How is the number e defined?

Mr. Cooper is making homework packets. He will put 6 worksheets in each packet. There are 18 worksheets. How many packets can he

ziro4ka [17]

6 times 18

Answer is 108

8 0

2 years ago

Please Helppppppppppp

sineoko [7]

Answer:

The width of the actual pool is 12 meters

Step-by-step explanation:

You can put Kiara's scale into a ratio or fraction to help visualize it better: 13/6. Now, let's put what we do know for the other ratio into a fraction. we know that the pool is 26 centimeters wide, and since the 26 matches with the 13, that goes in the numerater. since we don't know the denominator, we can just put 26/x (you can use whatever letter you like but I'll use x here)

So in order to get 13/6 to 26/x, we need to find how to figure out what we are doing to the original fraction. Since 13 x 2= 26, it looks like we are multiplying the ratios by 2. Now that we know that, we can multiply the denominator (6) by 2 to equal 12, our answer.

Hope this helps :)

8 0

2 years ago

These figures are similar. The area of one is given. Find the area of the other. PLZ HELP

stellarik [79]

Answer:

12 in squared

Step-by-step explanation:

Since the side on the bigger figure is 6 and the smaller figure‘s side is 3, that means the smaller figure is half the size of the big figure.

So.. 24 divided by 2 is 12

6 0

2 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER

mash [69]

Answer:

(-6,9,-3)

Step-by-step explanation:

-3x -y +z=6

-3x-y+3z =0

x-3z =3

Multiply the second equation by -1

-1 *(-3x-y+3z) =0*-1

3x +y -3z =0

Add this to the first equation

-3x -y +z=6

3x +y -3z =0

----------------------

0 + 0 + -2z = 6

Divide by -2

-2z/-2 = 6/-2

z = 6/-2

z=-3

Take the third equation to find x

x-3z=3

x-3(-3) = 3

x+9=3

Subtract 9 from each side

x+9-9 =3-9

x=-6

Now we need to find y

3x +y -3z =0

3(-6) +y -3(-3) =0

-18 +y +9=0

-9+y =0

Add 9 to each side

-9+9+y = 0+9

y=9

(-6,9,-3)

3 0

3 years ago

Read 2 more answers

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago