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blsea [12.9K]
3 years ago
9

Zach's office recycled a total of 6 kilograms of paper over 3 weeks. How many weeks will it take Zach's office to recycle a tota

l of 10 kilograms of paper? Assume the relationship is directly proportional.
Mathematics
2 answers:
Luba_88 [7]3 years ago
8 0
The answer to your question is five weeks
Korolek [52]3 years ago
3 0

Answer: 5 weeks.

Step-by-step explanation:

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F(n)=4n <br> g(n)=n2+2n <br> find f(g(-6))
ELEN [110]

Answer:

When f(n) = 4n and g(n) = n² + 2n, f(g(-6)) = 96.

Step-by-step explanation:

To evaluate f(g(-6)), first find g(-6).

g(n) = n² + 2n

Substitute value.

g(-6) = (-6)² + 2(-6)

Square -6. Remember that (-x)² = x²

g(-6) = 36 + 2(-6)

Multiply 2 and -6.

g(-6) = 36 - 12

Subtract 12 from 36.

g(-6) = 24.

Now knowing this, substitute that value into f(n).

f(g(-6)) = f(24)

f(n) = 4n

Substitute value.

f(24) = 4(24)

Multiply 4 and 24.

f(24) = 96.

8 0
3 years ago
PLEASS HELP!!
Vinvika [58]

Answer:

The answer is B

Step-by-step explanation:

By proportion , 3 : 2.91 = 2: x . then you write it in fraction

5 0
2 years ago
Element X decays radioactively with a half life of 9 minutes. If there are 890 grams of Element X, how long, to the nearest tent
elena-s [515]

Answer:

13.18651685

13.9

Explain:

890-238=652
652/890=0.7325842697 (Find out the percentage)

We know that 1 gram of Element X will be fully decayed in 18 minutes and half of it in 9 minutes.

0.7325842697x18=13.18651685

As you see 18/13.18651685 will equal to 18 (not exactly since it is rounded)

And of course we need to round it to nearest tenth. So the answer would be 13.9!

3 0
2 years ago
Plz help me some plz, i posted this question 1000 times plz some one answer, thank you so much plzz show all the working i reall
Dima020 [189]

Answer:

sorry man I really don't know hopefully you can find an answer

3 0
3 years ago
The line 5x – 5y = 2 intersects the curve x2y – 5x + y + 2 = 0 at
inna [77]

Answer:

(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve at each point of intersection are;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28

Step-by-step explanation:

The equations of the lines are;

5·x - 5·y = 2......(1)

x²·y - 5·x + y + 2 = 0.......(2)

Making y the subject of equation (1) gives;

5·y = 5·x - 2

y = (5·x - 2)/5

Making y the subject of equation (2) gives;

y·(x² + 1) - 5·x + 2 = 0

y = (5·x - 2)/(x² + 1)

Therefore, at the point the two lines intersect their coordinates are equal thus we have;

y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)

Which gives;

\dfrac{5 \cdot x - 2}{5} = \dfrac{5 \cdot x - 2}{x^2 + 1}

Therefore, 5 = x² + 1

x² = 5 - 1 = 4

x = √4 = 2

Which is an indication that the x-coordinate is equal to 2

The y-coordinate is therefore;

y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5

The coordinates of the points of intersection = (2, 8/5}

Cross multiplying the following equation

Substituting the value for y in equation (2) with (5·x - 2)/5 gives;

\dfrac{5 \cdot x^3 - 2 \cdot x^2 - 20 \cdot x + 8}{5} = 0

Therefore;

5·x³ - 2·x² - 20·x + 8 = 0

(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8

Therefore, we have;

x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²

5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8

∴ c = 8/(-2) = -4

2·b + c = - 20

b = -16/2 = -8

Therefore;

(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)

(x - 2)×(5·x² + 8·x - 4) = 0

5·x² + 8·x - 4 = 0

x² + 8/5·x - 4/5  = 0

(x + 4/5)² - (4/5)² - 4/5 = 0

(x + 4/5)² = 36/25

x + 4/5 = ±6/5

x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2

Hence the three x-coordinates are

x = 2, x = - 2, and x = 2/5

The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0

The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve, \dfrac{\mathrm{d} y}{\mathrm{d} x}, is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)

Therefore, we have;

\dfrac{\mathrm{d} y}{\mathrm{d} x}= \dfrac{\mathrm{d} \left (\dfrac{5 \cdot x - 2}{x^2 + 1}  \right )}{\mathrm{d} x} = \dfrac{5\cdot \left ( x^{2} +1\right )-\left ( 5\cdot x-2 \right )\cdot 2\cdot x}{\left (x^2 + 1 ^{2} \right )}.......(3)

Which gives by plugging in the value of x in the slope equation;

At x = -2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.92

At x = 2/5, \dfrac{\mathrm{d} y}{\mathrm{d} x} = 4.3

At x = 2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.28

Therefore;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28.

7 0
4 years ago
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