Evaluate 16 3/4 nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

Please answer the top half ASAP I need it in 2h please

NeTakaya

Answer:

4/7
6/11
12/17
19/30
10/17
7/15
11/28
8/21
7/3
23/49
3/7
1/2
4/7
5/22
3/7
3/4
1/3
1/10
3/8
1/2
3/8
4/7
65/168
8/21
40/63
4/3
1/2
9/7
9/5
21/44

4 0

3 years ago

On a coordinate plane, 2 lines intersect around (0.4, 2.8). The graph on the left represents a system of equations. How can the

kompoz [17]

Answer:

How can the solution be located?

find the intersection

Which integers is the x-coordinate between?

0 and 1

Which integers is the y-coordinate between?

2 and 3

What is the solution approximated to the tenths place?

(0.4,2.8)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Its for science but its got math involved its got to do with speed distance and time

KIM [24]

distance:1, 1, 1.3, 1.5, 2, 2.3, 2.5, 3 ,3.2, 4

time. :1, 2, 3 , 4 , 5 , 6 , 7 , 8 ,. 9 ,10

distance:1.5, 2, 3.2, 4.2, 5, 6, 7, 8

time. : 1 , 2, 3 , 4. , 5, 6, 7, 8

<h3>at 5 seconds which travelled the greatest distance : snail 2</h3>

<h3>which travelled a greater distance:snail 2</h3>

<h3>what is the speed of snail 1 at 8 second :</h3>

distance = 3. Time = 8

speed = distance

time

speed = 8

3

speed =2.66

approximately = 2.7. or 3

7 0

3 years ago

A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs

LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

$h=\frac{400}{\pi r^2}$

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: $h=\frac{400}{\pi r^2}$

Therefore:

$C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})$

$C(r)=0.12\pi r^2+\frac{24}{r}$

$C(r)=\frac{0.12\pi r^3+24}{r}$

The minimum cost occurs when the derivative of the Cost =0.

$C^{'}(r)=\frac{6\pi r^3-600}{25r^2}$

$6\pi r^3-600=0$

$6\pi r^3=600$

$\pi r^3=100$

$r^3=\frac{100}{\pi}$

$r^3=31.83$

r=3.17 cm

Recall that:

$h=\frac{400}{\pi r^2}$

$h=\frac{400}{\pi *3.17^2}$

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0

3 years ago

Maths question trigonometry help pleassssee

r-ruslan [8.4K]

Answer:

x=4.5

Step-by-step explanation:

Using sohcahtoa we know tangent is equal to (length of side opposite to the angle ) / (lenth of side adjacent to the angle)

therefore: tan (theta) = (x-3) / (x) = 1 / 3

now cross multiply

3(x-3) = x

now solve for x

3x - 9 = x

-9= -2x

x= 4.5

7 0

2 years ago