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disa [49]
3 years ago
13

A painter spends 3 hours working on a painting. A sculptor spends 2 2/3 as long working on a sculpture. How long does the sculpt

or work?
Mathematics
2 answers:
valentina_108 [34]3 years ago
4 0
He would have worked 8 hours.
kompoz [17]3 years ago
3 0
I believe its 8 hours hat he sculped
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5x+18/8 = x/4 <br> Solve for x
ollegr [7]

Answer: x= -9/19

Step-by-step explanation:

5x+18/8=x/4

5x+ 9/4=x/4

5x+9/4-9/4=x/4-9/4

5x= x-9/4

5x(4)=x-9/4(4)

20x=x-9

20x-x=x-9-x

19x=-9

19x/19=-9/19

x=-9/19

8 0
3 years ago
What are the next few terms after the following sequence? 1, 1/2, 1/3, 1/4
DerKrebs [107]

Answer:

1/5, 1/6, 1/7, 1/8,1/9, 1/10,1/11,1/12

3 0
2 years ago
5≤3y+8&lt;17 solve and graph
Scorpion4ik [409]
The answer to your question is -1 < y < 3
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4 0
2 years ago
What are the roots of the quadratic equation below 2x^2-5x-1=0
PilotLPTM [1.2K]

Answer:

roots=(5±sqrt(33))/4

Step-by-step explanation:

roots=(-b±sqrt(b^2-4ac))/2a

roots=(-(-5)±sqrt(25+8))/(2*2)

roots=(5±sqrt(33))/4

roots=(5+sqrt(33))/4 and (5-sqrt(33))/4

7 0
2 years ago
Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
6 0
3 years ago
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