Plz Help NO LINKS NO LINKS

nexus9112 [7]

First multiply 9 by 60 to find feet/minute: 9x60=540. Multiply that answer by 60 to get feet/hour: 540x60=32400. Divide that by 5,280 to get miles/hour: 32400/5280=6.14 roughly :)

4 0

2 years ago

how do you solve a system of equations approximately given tables without using equations or graphing?

solong [7]

system of equations means that we have a given number of equations with the same solutions.

If you only have tables, this means that you need to have one table for each equation:

For example, if you are working only with two variables, x and y, in those tables you can see the pints (x, y) that belong to each equation.

Now, a point (x, y) will be a solution of the system of equations only if it belongs to the data table for each equation

This would mean that if you graph those data sets, the graphs will intersect at the point (x, y) that belongs to all the tables of data.

another way may be using the data in the tables to construct the equations, but you said that you only want to use the tables, so this method can be discarded.

7 0

2 years ago

Check the picture of the necessary information

denis-greek [22]

The probability would be out of 100,
in this case if it refers to the data from 2009 it would be 39:61

4 0

1 year ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

2 years ago

Can someone do this for me or help me

Elden [556K]

JM=12
JL= 14
MN=?
MK=?

VT= 11
UV= 9
RS=?
ST=?

GF=23
HF=20
GH=?
GE=?

M<1=?
M<2=?
M<3=?
M<4=?
M<5=?
M<6=?
M<7=?
M<8 = 90 degrees

WXZ = 34 degrees
WVZ=90 degrees
ZYW= 56 degrees

These are the only answers I knew, I’m sorry I couldn’t find the rest. If I do find more answers, I’ll comment them.

5 0

2 years ago