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2 years ago

Solve the equation. Show all your work. Put your final answer in the blank.) 12x + 9 1 = 12

Mathematics

2 answers:

qaws [65]2 years ago

6 0

Answer:

x=(-79/12) or -6 7/12

Step-by-step explanation:

12x+91=12

-91 -91

12x=-79

(12x/12)=(-79/12)

x=(-79/12) or -6 7/12

Rufina [12.5K]2 years ago

3 0

Hope this help u and good luck

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Students went on a picnic. They recorded a list of the number of animals

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5 0

3 years ago

A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, positive 1, positive 2. The

S_A_V [24]

The rate of change of the linear function represented by the table is 3

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

The standard linear equation is in the form:

y = mx + b

Where m is the rate of change and b is the y intercept.

From the table, using the point (0, -7) and (2, -1):

m = [-1 - (-7)] / (2 - 0) = 3

The rate of change of the linear function represented by the table is 3

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4 0

1 year 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