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3 years ago

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Choose = , < , or > so that each statement is true. When x = 0, y1 y2 When x = 5, y2 y3 Choose y1, y2, and y3 so that the

statement is true. When x = -1, < <

1 answer:

serg [7]3 years ago

3 0

< because you wouldnt need >

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Which expressions are equivalent to the given expression?

gizmo_the_mogwai [7]

Answer: Choice C. $\frac{1}{x^{2}y^{5} }$ and Choice E. $x^{-2} y^{-5}$

Step-by-step explanation:

Algebraic exponents.

5 0

3 years ago

Read 2 more answers

Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|

Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0

3 years ago

Factor completely

saul85 [17]

Answer:

c4-2401=0 because o is not a prime number

8 0

3 years ago

Given circle O. Find all the missing values

Norma-Jean [14]

Step-by-step explanation:

Do you still need the answer to this?

3 0

3 years ago

How many solutions does the system of equations have? Pls help :(

goldfiish [28.3K]

Given the system of equations below:

$\large{ \begin{cases} y = 2x + 1 \\ - 4x + 2y = 2 \end{cases}}$

The first equation is y-isolated so we can substitute in the second equation.

$\large{ - 4x + 2(2x + 1) = 2}$

Use the distribution property to expand in and simplify.

$\large{ - 4x + 4x + 2 = 2} \\ \large{0 + 2 = 2 \longrightarrow 2 = 2}$

The another method is to divide the second equation by 2.

$\large{ \frac{ - 4x}{2} + \frac{2y}{2} = \frac{2}{2} } \\ \large{ - 2x + y = 1}$

Arrange in the form of y = mx+b.

$\large{y = 1 + 2x \longrightarrow y = 2x + 1}$

When we finally arrange, compare the equation to the first equation. Both equations are the same which mean that both graphs are also same and intersect each others infinitely.

For more information, when the both sides are equal for equation - the answer would be infinitely many. If both sides aren't equal (0 = 4 for example) - the answer would be none. If the equation can be solved for a variable then it'd be one solution.

Answer

Infinitely Many

Hope this helps. Let me know if you have any doubts!

3 0

3 years ago