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ch4aika [34]
3 years ago
15

I need to know how to graph this inequality

Mathematics
1 answer:
Marizza181 [45]3 years ago
3 0
Graph it like a normal line, then shade in everything above the line to show the greater than part
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I need to know what’s 67x4
Simora [160]

Answer:

268

Step-by-step explanation:

please give brainliest

3 0
3 years ago
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Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
4 years ago
Select the correct answer from the drop-down menu.
ELEN [110]

Missing term = –2xy

Solution:

Let us first find the quotient of -8x^2y^3 \div xy.

-8x^2y^3 \div xy=\frac{-8x^2y^3 }{xy}

                    =\frac{-8\times x\times x\times y\times y\times y}{xy}

Taking common term xy outside in the numerator.

                    =\frac{xy(-8\times x\times y\times y)}{xy}

Both xy in the numerator and denominator are cancelled.

                    =-8xy^2

Thus, the quotient of -8x^2y^3 \div xy is -8xy^2.

Given the quotient of -8x^2y^3 \div xy is same as the product  of 4xy and ____.

-8xy^2=4xy × missing term

Divide both sides by 4xy, we get

⇒ missing term = \frac{-8xy^2}{4xy}

Cancel the common terms in both numerator and denominator.

⇒ missing term = –2xy

Hence the missing term of the product is –2xy.

4 0
3 years ago
Describe what is being done in each step while solving the equation. Match each step with the correct description. HINT: Start w
Licemer1 [7]
1. Original equation
2.Distributive property
3. Add 6x to each side
4. Subtract 2 from each side
5. Divide each side by 11
5 0
3 years ago
What is the slope of a line that is perpendicular to the line y = 1?
ValentinkaMS [17]
Y = 1
Using y = mx + c.
Compare to y = 1,    y = 0x + 1 ,
We can see that the slope m = 0 and the vertical intercept, c = 1.
For the line perpendicular to  y = 1
Condition for perpendicularity  m₁m₂ = -1
m₁ = 0,  m₂ = ?
0*m₂ = -1
m₂ = -1/0 = Negative Infinite or Infinite 
Slope of line perpendicular to y = 1, is  = Infinite. 
3 0
3 years ago
Read 2 more answers
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