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icang [17]
3 years ago
13

Does the point (6, 0) satisfy the equation y = x2?

Mathematics
2 answers:
Nataly_w [17]3 years ago
7 0

Answer:

No, point (6, 0) is not on the equation.

Step-by-step explanation:

To do this question the easiest way, you would use your scientific/graphing calculator and type in your equation. But you can do this with your mind.

Since the equation y = x^2 does not have any number in it (such as m = slope) it does not start anywhere. You will put it in the origin which is (0, 0) from there, you can tell that the equation will not reach (6, 0), but only (1, 1).

shtirl [24]3 years ago
6 0

Replace x in the equation with the x value of the point (6) and solve. If it equals the y value (0) it is a solution if it noes not equal (0) it is not a solution.

Y = 6^2 = 36

36 is not 0 so (6,0) is not a solution

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Solve for x, round answer to nearest thousandth<br><br> 22^3x+2=114
vazorg [7]

Answer:

x=28/363 or x=0.077134

Step-by-step explanation:

484×3x=114-2

1452x=112

x=112/1452

x=28/363

7 0
3 years ago
Need help on angles of polygons please
aleksandrvk [35]

Answer:

x+87+92+105+135=540(sum of interior angle of pentagon)

x+419=540

x=540-419

x=121

Step-by-step explanation:

4 0
3 years ago
sophie was 28 years old when daughter Nichole was born. In how many years will Sophie be two years less than still times Nichole
Kamila [148]
It will take 6 years for that to happen. 

You can find this by setting Nichole's age as x and Sophie's age as x + 28 since she will always be 28 years older. 

Now we are looking for a time that Sophie's age is equal to 2 less than 6 times Nichole's age. So we can set up the below equation. 

6(x) - 2 = x + 28
6x - 2 = x + 28
5x - 2 = 28
5x = 30
x = 6

This let's us know that it needs to be when Nichole is 6, which is 6 years in the future. 
6 0
3 years ago
Find the nth Maclaurin polynomial for the function. f(x) = sec(x), n = 2 P_2(x) =
Alika [10]

Answer:

\mathbf{P_2(x) = 1+\dfrac{x^2}{2}}

Step-by-step explanation:

Given that:

f(x) = sec (x)  , n = 2

Where are to find P_2(x)

Suppose ; f(x) = sec (x)  , n = 2

then

f(0) = sec (0) = 1

f'(x) = sec (x)* tan (x)|_{x=0} = 0

f''(x) = sec (x)*tan ^2(x)+ sec (x) * sec^2(x)

f''(x) = sec (x)*tan ^2(x)|_{x=0} + sec^3(x)

f''(x) = 0 + sec^3(0)

f''(x) = 1

f(x) = f(0) + \dfrac{f'(0)x}{1!}+ \dfrac{f''(0)x^2}{2!}+  \dfrac{f'''(0)x^3}{3!}+...

f(x) = 1 + \dfrac{0}{1!}x+ \dfrac{x^2}{2!}+...

f(x) = 1 + \dfrac{x^2}{2}+...

since order n =2

\mathbf{P_2(x) = 1+\dfrac{x^2}{2}}

8 0
3 years ago
What is 2x-3+8 simplified
zavuch27 [327]

Answer: 2x+5

Step-by-step explanation:

Add  − 3  and  8 .

8 0
4 years ago
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