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Readme [11.4K]
2 years ago
13

[tex]40-23\frac{6}{7}

Mathematics
2 answers:
pshichka [43]2 years ago
5 0

Answer:

Step-by-step explanation:

40 minus 23 and six-sevenths

A. 17 and one-seventh

B. 17 and six-sevenths

C. 16 and one-seventh

D. 16 and six-sevenths

adelina 88 [10]2 years ago
3 0

Answer:

<h3>What kind of questions is thisss so what's the point0385729484 </h3>
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1. If h(x) = 4x - 2, then which of the following is the solution to h(x) = 3 ?
luda_lava [24]

Answer:

10

Step-by-step explanation:

h(x)=4x-2                         h(x)=3

h(3)=4(3)-2                     substitute

h(3)=12-2                        multiply

=10                                subtract

8 0
2 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
Solve the equation.<br><br> y + 6 = –3y + 26
Sergeeva-Olga [200]
     y + 6 = -3y + 26
+ 3y        + 3y 
   4y + 6 = 26
         - 6   - 6
         4y = 20
          4      4
           y = 5
3 0
3 years ago
Read 2 more answers
E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis .Find the coordinates of H and G. Please need answe
timurjin [86]

Answer:

coordinates of H = (1, 0)

Coordinates of G = ( - 3.6, -2) or (5.6, -2)

Step-by-step explanation:

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis.

Let the coordinates of H is (x, 0) and G is (a, b).

The length of side EF is

EF = \sqrt{(5 -2)^2 + (3 +1)^2} = 5

So,

EH = \sqrt{(5 -x)^2 + (3 -0)^2} = 5\\\\(5 -x)^2+ 9 = 25\\\\5 - x = 4\\\\x = 1

And

GH = \sqrt{(a -x)^2+ b^2} = 5\\\\(a -1)^2+ b^2 = 25\\\\a^2 + b^2 + 1 - 2 a = 25\\\\a^2 + b^2 - 2a = 24 .... (1)

Now

FG = \sqrt{(a -2)^2+ (b +1)^2} = 5\\\\(a -2)^2+ (b+1)^2 = 25\\\\a^2 + b^2 + 2b - 2 a = 20\\   ..... (2)

Solving (1) and (2)

b = - 2 ,

a = \frac{2 \pm\sqrt{4 + 80}}{2}\\\\a = \frac{2 \pm 9.2}{2}\\\\a = - 3.6, 5.6

3 0
3 years ago
For a certain population, the regression equation to predict salary (in dollars) from education (in years) is y=2530x + 5200. Wh
timofeeve [1]

Answer:

2530 has no units

Step-by-step explanation:

In order to understand the units from a linear equation we need to understand the general equation of a line which is:

y=mx+b where:

m=slope of the line

b=y-intercept.

Comparing the given equation with the general line equation, we noticed that 2530 represents the slope of the line.

Since the slope can be obtained by:

m=(y2-y1)/(x2-x1) whatever the units are, the slope is  dimensionless, which means that 2530 has no units.

3 0
3 years ago
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