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suter [353]
3 years ago
6

When you graph a system of equations what are the three possible outcomes?

Mathematics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

i think its a beaner

Step-by-step explanation:

You might be interested in
8 grade math dont really uderstand it so please explain also
pishuonlain [190]

Answer: Girls weekly: 25; Girls never:40; boys yearly: 25; boys never: 20; boys total:80 monthly total: 30

Step-by-step explanation:

You have to figure it out by adding or subtracting where ever it needs to be like in girls weekly you have to subtract the weekly total and the boys weekly to figure it out.

5 0
3 years ago
In parallelogram HIJK, the measure of angle H is 45 degrees. 1.Find the measure of J and explain how you know
s2008m [1.1K]

Answer:

45°

Step-by-step explanation:

A parallelogram is a polygon with 4 sides and 4 angles (that is a quadrilateral).

For a parallelogram ABCD, Opposite sides and opposite angles are equal (i.e. BC = AD and ∠A = ∠C).

Consecutive angles are supplementary that is ∠A + ∠B = 180°.

The diagonals of a parallelogram bisect each other also separating the parallelogram into two congruent triangles.

In parallelogram HIJK, ∠H and ∠J are opposite angles. hence:

∠H = ∠J (opposite angles of a parallelogram are equal)

∠J = 45°

8 0
3 years ago
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
Is h(x) = 6 a linear function and why
Andrei [34K]

Answer:

yes, absolutely. specifically it's a horizontal line at 6.

6 0
2 years ago
If f(x) = x + 5, what does f(8) represent? (1 point) Group of answer choices The value of x when (x + 5) = 8 The value of 8f(x)
baherus [9]

Answer:

a

Step-by-step explanation: thank me later

3 0
3 years ago
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