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3 years ago

What is the width of a rectangle with length 14 cm and area 161cm squared 2

1 answer:

3 0

Length • Width = area. Length is 14. So set it up like an equation. W • 14 = 161. This simplifies to 14W= 161. Divide both sides by 14 to get the variable alone. This gets you 11.5. So the width is 11.5. Hope this helps

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A farmer can buy two types of plant food, mix A and mix B. Each cubic yard of mix A contains 35 pounds of phosphoric acid, 15

Nadusha1986 [10]

Answer:

The answer is below

Step-by-step explanation:

Given that the minimum monthly requirements are 840 pounds of phosphoric acid, hence:

35 pounds of phosphoric acid mix A + 14 pounds of phosphoric acid mix B ≥ 840 pounds of phosphoric acid

35x + 14y ≥ 840

The minimum monthly requirements are 540 pounds of nitrogen, hence:

15 pounds of nitrogen mix A + 18 pounds of nitrogen mix B ≥ 540 pounds of nitrogen

15x + 18y ≥ 540

The minimum monthly requirements are 768 pounds of potash, hence:

16 pounds of potash mix A + 64 pounds of potash mix B ≥ 768 pounds of potash

16x + 64y ≥ 768

Also, x ≥ 0 and y ≥0.

Therefore the equations are:

35x + 14y ≥ 840

15x + 18y ≥ 540

16x + 64y ≥ 768

x ≥ 0 and y ≥ 0

Plotting with geogebra gives the graph below.

The point are (0,60), (48,0), (18,15), (32,4)

7 0

2 years ago

Square root of 80 to the power of 2

Alex Ar [27]

80 would be your answer for this question

3 0

3 years ago

Read 2 more answers

One kind of hard candy sells for $.89 per kilogram another sells for $1.10 per kilogram how many kilograms of each kind a need t

Furkat [3]

Answer:

20kg of $0.89 candy

10kg of $1.10 candy

Step-by-step explanation:

Candy 1 = 0.89 per kg

Candy 2 = 1.10 per kg

Total kilogram, kg = 30

Let candy 1 = x ; candy 2 = (30 - x) ;

0.89x + 1.10(30 - x) = 0.96(30)

0.89x + 33 - 1.10x = 28.8

0.89x - 1.10x = 28.8 - 33

-0.21x = - 4.2

x = 4.2 / 0.21

x = 20

20kg of $0.89 candy

(30 - x) = (30 - 20) = 10kg

10kg of $1.10 candy

4 0

2 years ago

What is the inverse of the function g(x) = x^3/8 + 16 ?

lesya692 [45]

Answer:

$\hookrightarrow \: { \rm{f(x) = \frac{ {x}^{3} }{8} + 16}} \\$

• let f(x) be m:

${ \rm{m = \frac{ {x}^{3} }{8} + 16}} \\$

• make x the subject of the function:

${ \rm{8m = {x}^{3} + 128}} \\ \\ { \rm{ {x}^{3} = 8m - 128 }} \\ \\ { \rm{ {x}^{3} = 8(m - 16) }} \\ \\ { \rm{x = \sqrt[3]{8} \times \sqrt[3]{(m - 16)} }} \\ \\ { \rm{x = 2 \sqrt[3]{(m - 16)} }}$

• therefore:

${ \boxed{ \rm{ {f}^{ - 1} (x) = 2 {(m - 16)}^{ \frac{1}{3} } }} }\\$

6 0

3 years ago

What is dislocation

Sati [7]

Answer:

The distance between the starting point and the end point

Step-by-step explanation:

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3 years ago