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RUDIKE [14]
2 years ago
10

PLEASE ANSWER ASAP

Mathematics
1 answer:
professor190 [17]2 years ago
7 0

Answer:

hmmm I think u need to talk to ur teacher

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I need 2-4 please & show step by step explanation ( show the work ) please!
Likurg_2 [28]

Answer:

i don't know go on your browser

Step-by-step explanation:

3 0
2 years ago
use the plot of the concentration of cu(no3)3 versus absorbance provided. if the absorbance of an unknown was determined to be 0
Helen [10]

Using the linear regression equation, the concentration of the unknown solution is 0.2161 M.

Linear regression describes the relationship of two variables. It may not be exact but it is the line that best fit the data. The equation for a linear regression is in the form y = bx + a, where x and y are the two variables.

If the absorbance of an unknown was determined to be 0.67 absorbance units, using the linear regression equation provided from the plot, substitute the value of absorbance to the variable y and solve for the value of x or the concentration.

y = 3.8674x - 0.1657

0.67 = 3.8674x - 0.1657

3.8674x = 0.67 + 0.1657

3.8674x = 0.8357

x = 0.2161

Hence, the concentration is 0.2161 M.

Learn more about linear regression here: brainly.com/question/25311696

#SPJ4

6 0
1 year ago
3/8 + 4/8 = ? please help
maksim [4K]
The answer is going to be 7/8. hope that helped
4 0
3 years ago
Read 2 more answers
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
3 years ago
I need help with 17 please
stira [4]
Move the triangle, 12 units right and 4 units bottom. Then, rotate it by 180 degrees. Transformation would be done!

Hope this helps!
7 0
3 years ago
Read 2 more answers
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