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Alja [10]
3 years ago
13

Is this a direct variation? 4y=x

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
5 0
Google it is easy
sorry
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What is the answer??
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The answer is 30, assuming it is a cube, 
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Try this trick out on a friend. Tell your friend to place a dime in one hand and a penny in the other hand. Explain that you can
lyudmila [28]

Answer:

does it work

Step-by-step explanation:

7 0
3 years ago
Help me with this please #17
jasenka [17]

9514 1404 393

Answer:

  C. symmetric with respect to the origin

Step-by-step explanation:

A graph must pass the vertical line test to be called the graph of a function. No function will ever be symmetrical with respect to the x-axis, because that would mean it fails the vertical line test.

An even function is symmetrical about the y-axis. An even function has the characteristic that ...

  f(-x) = f(x) . . . . an even function

An odd function is symmetrical about the origin. An odd function has the characteristic that ...

  f(-x) = -f(x)

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The given function is f(x) = -1/x. Then the value of f(-x) is ...

  f(-x) = -1/-x = 1/x = -f(x)

The given function is an odd function, so is symmetrical about the origin.

5 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
3 years ago
2.9 x 10^5 + 8.7 x 10^5
LiRa [457]

Answer:

So this is scientific notation what you do is

2.9x10^5 so you put however many zeros the is powered to 10. See there is a 5 It has to be behind or in front of the decimal you might be saying how do you know if its behind or in front well negative is in front and positive is behind .

<u><em>290,000 </em></u>there was a number already behind the decimal so there you go 5 number behind the decimal.  

Same for the second one

<u><em>290,000</em></u><em>  </em><u><em>870,000</em></u> is your answers so now you add hem together

<em><u>1,160,000 is your final answer </u></em>

Step-by-step explanation:

3 0
3 years ago
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