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GuDViN [60]
4 years ago
7

Identify the oblique asymptote of f(x)=2x2-5x+2 over x-3

Mathematics
1 answer:
Katyanochek1 [597]4 years ago
7 0
D) No oblique asymptote
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Raj and Aditi like to play chess against each other. Aditi has won 12 of the 20 games so far. They decide to play more games. Ho
Over [174]
12 + x = 0.80(20+x)
12 + x = 16 + 0.80x
12 = 16 -0.2x
-4 = -0.2x

x = -4/-0.2 = 20
 they need to win 20 more games


4 0
3 years ago
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Translate this sentence into an equation.
vodomira [7]

Answer:

3x-x+2=4

Step-by-step explanation:

4 0
3 years ago
You are given an equation of a line and a point. use substitution to determine whether the point is on the line.
Anon25 [30]

Answer:

(1,10) That is the slope, if there is another line then the answer will be different!

Step-by-step explanation:

This is the slope-intercept formula, (y=mx+b)

Start by finding the y-intercept which is Positive 7 or plus 7 you should find it on the y-axis on the positive side.

Then you will need to identify the slope, the slope, in this case is 3, which needs to be in rise over run formula so 3 over 1.

To find a slope you must start at the y-intercept in your case is 7, then you must go up 3 and right 1.

Then identify the x and y axis.

If this is not the answer please provide this equation and another one to get them intersecting to get one point.

4 0
3 years ago
Please find the surface area of the sphere. Round your answer to the nearest hundredth.
Elenna [48]

Surface area = \frac{4}{3} \pi r^{3} = \frac{4}{3} x 3.14 x 3 x 3 x 3 = 3.14 x 4 x 3 x 3 = 113.04 yd^2 = approx. 113.1 yd^2

4 0
4 years ago
Solve the equation after making an appropriate substitution. X^6-98x^3=3375
Leni [432]
Rearrange x^6-98x^3=3375 to give

x^6-98x^3-3375=0

Let x^3 be a variable 'p' and so we can write x^6 as x^6=(x^3)(x^3)=(p)(p)= p^2

Rewrite the equation in terms of 'p'

p^2-98p-3375 = 0

where a=1, b=-98, c=-3375

using the quadratic formula \frac{-b+ \sqrt{b^2-4ac} }{2a} and subsitute the value of a, b, c

p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125

p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27

There are two value of p; 125 and -27 

Now we find the value of x

Earlier we substitute x^3 for p and x^6 for p^2

When p=125, x= \sqrt[3]{125}=5
When p = -27, x= \sqrt[3]{-27}=-3

So the final answer is the two values of x; 

x = 5 OR x = -3
3 0
3 years ago
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