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swat32
1 year ago
11

Find the slope of this line.

Mathematics
1 answer:
natulia [17]1 year ago
4 0

Answer:

4/5

Step-by-step explanation:

SImply use slope formula. We see the graph has defined points at (5, 4), and (-5, -4).

m=(Y2-Y1)/(X2-X1)

(4-(-4))/(5-(-5))=8/10

Simplify 8/10

=4/5

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Solve the inequality.<br> Graph the solution on a number line.
Lina20 [59]

Answer:

x greater than or equal to -3

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In order to solve the system of equations below, Harvey multiplies each equation by a constant to eliminate the x terms.
Karo-lina-s [1.5K]

Answer:

<u>The correct answer is A. 14 x + 6 y = 10. Negative 14 x minus 35 y = 77</u>

Step-by-step explanation:

Let's find out the constants to eliminate the x terms

7 x + 3 y = 5

2 x + 5 y = negative 11

Let's multiply the first equation by 2 and the second one by -7, this way:

2 * (7 x + 3 y = 5)

-7 * (2 x + 5 y = negative 11)

14x + 6y = 10

-14x - 35y = 77

<u>The correct answer is A. 14 x + 6 y = 10. Negative 14 x minus 35 y = 77</u>

5 0
3 years ago
Read 2 more answers
A triangle is cut from a<br> rectangle. Find the area<br> of the shaded region.
KATRIN_1 [288]

The area of rectangle:

11 ft · 6 ft = 66 ft²

The area of triangle:

6 ft · 4 ft/2 = 6 ft · 2 ft = 12 ft²

The area of the shaded region:

66 ft² - 12 ft² = 54 ft²

4 0
3 years ago
Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an
Andrej [43]

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as V=g(x), where g(x)=x(6-2x)(8-2x) and 0\le x\le3.

Expand the function to obtain:

g(x)=4x^3-28x^2+48x

Differentiate  wrt  x to obtain:

g'(x)=12x^2-56x+48

To find the point where the maximum value occurs, we solve

g'(x)=0

\implies 12x^2-56x+48=0

\implies x=1.13,x=3.54

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

g''(x)=24x-56, g''(1.13)=24(1.13)-56=-28.88\:

This means the maximum volume occurs at x=1.13.

Substitute x=1.13 into g(x)=x(6-2x)(8-2x) to get the maximum volume.

g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

6 0
3 years ago
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