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maks197457 [2]
3 years ago
12

Help with answering quoin photo

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0

K=24

S=48

SS=72

all together: 144

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How many pounds are there in a 72-ounce package of ground beef?(1 pound=16 ounces)
UkoKoshka [18]

Answer:

4.5 pounds of ground beef

3 0
3 years ago
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BRAINLIESTTTTTT ASAPPPP<br> PLEASE ANSWER MY QUESTION
yawa3891 [41]
The red line is represented by y ≥ -3x + 3.

The blue line is represented by y < 3/2x - 6.

A) In this system of inequalities, there is one solid line and one dashed line.

The solid line is the inequality y ≥ -3x + 3 because it has the symbol greater than or equal to. The "equal to" part of the symbol is important because it means that any coordinate located on the solid line is a solution.

However, the inequality y < 3/2x - 6 is a dashed line because it does not have "equal to." This makes the inequality a dashed line which means that all coordinates on it will not be a solution to the inequality.

In the inequality y ≥ -3x + 3, the shading is above the line. This means that all coordinates within the line and shaded area above the line will be a solution to the inequality.

In the inequality y < 3/2x - 6, the shading is below the line. This means that all coordinates within the shared area of the line will be a solution to the inequality. However, since y is not less than or EQUAL TO, all coordinates within the dashed line is not a solution to the inequality.

The solution area, where a coordinate is a solution to
both inequalities, is where the shaded areas of both inequalities overlap.

B) The point (-6, 3) is a not solution to both of the inequalities because it is not within the shared areas of the two inequalities.

Substitute the coordinate (-6, 3) in the inequality y < 3/2x - 6.

y < 3/2x - 6

3 < 3/2(-6) - 6

3 < -9 - 6

3 < -15

This is not true because a negative number cannot be greater than a positive number. Therefore, this proves that the coordinate (-6, 3) is not a solution for the inequality y < 3/2x - 6.

Check if the coordinate fits for the other inequality.

y ≥ -3x + 3

3 ≥ -3(-6) + 3

3 ≥ 18 + 3

3 ≥ 21

This is also not true because 3 is not greater than or equal to 21. Therefore, this coordinate is not a solution to the inequality y ≥ -3x + 3.

The coordinate (-6, 3) is not a solution for the system of inequalities because it is not a solution for both inequalities.

7 0
3 years ago
Slope is 2 and (−1, 6) is on the line; point-slope form
MissTica

Answer:

y-6=2(x-(-1))

Or simplified:

y-6=2(x+1)

Step-by-step explanation:

use the formula: y-y1=m(x-x1)

8 0
3 years ago
The first derivative of x^2+2y^2=16 is -x/(2y), the second is -(2y^2+x^2)/(4y^3). Find the third implicit derivative of x^2+2y^2
gtnhenbr [62]

Answer:

d³y/dx³ = (-2xy² − 3x³ − 4xy²) / (8y⁵)

Step-by-step explanation:

d²y/dx² = (-2y² − x²) / (4y³)

Take the derivative (use quotient rule and chain rule):

d³y/dx³ = [ (4y³) (-4y dy/dx − 2x) − (-2y² − x²) (12y² dy/dx) ] / (4y³)²

d³y/dx³ = [ (-16y⁴ dy/dx − 8xy³ − (-24y⁴ dy/dx − 12x²y² dy/dx) ] / (16y⁶)

d³y/dx³ = (-16y⁴ dy/dx − 8xy³ + 24y⁴ dy/dx + 12x²y² dy/dx) / (16y⁶)

d³y/dx³ = ((8y⁴ + 12x²y²) dy/dx − 8xy³) / (16y⁶)

d³y/dx³ = ((2y² + 3x²) dy/dx − 2xy) / (4y⁴)

Substitute:

d³y/dx³ = ((2y² + 3x²) (-x / (2y)) − 2xy) / (4y⁴)

d³y/dx³ = ((2y² + 3x²) (-x) − 4xy²) / (8y⁵)

d³y/dx³ = (-2xy² − 3x³ − 4xy²) / (8y⁵)

8 0
3 years ago
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Need help with this work sheet
Lesechka [4]

Answer:

its alot to explain but i will try to make it as simple as possible

Step-by-step explanation:

your first goal is to make each problem into the form ax^2+bx+c=0

number 1, 2, 7 and 8 is already done for you

now all you have to do is plug in each number in the standard form into the quadtratic formula.

basically at this point you can just use your calculator to do the rest of the work. dont forget parentheses so it doesnt get confused...

or you can perform the algebraic work.. its all just a matter of plugging in the right numbers into the quadratic formula...

cant really do the work for you since im on my phone. but yeah all you need to do step one is transform each problem into ax^2+bx+c=0 form

then step 2, plug in each number in to the quadtratic formula. from there calculate using basic algebraic rules

5 0
3 years ago
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