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torisob [31]
3 years ago
5

Consider the quadratic function f(x) = x2 – 8x – 4. What is the value of the leading coefficient?

Mathematics
1 answer:
Tatiana [17]3 years ago
5 0
D
there is always a 1 infront of a variable but since x2 has the largest degree it gives it priority over the others so what ever numbe in front of it is the leading coefficient
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Please I need help!​
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Answer:

M and P

Step-by-step explanation:

they do not cross each other

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3 years ago
Water flows at a constant rate out of a faucet. Suppose the volume of water that comes out in three minutes is 10.5 gallons. How
Setler [38]

Answer:

3.5 Gallons Per Minute or 3.5m

Step-by-step explanation:

Since there is 10.5 Gallons in 3 minutes, do

10.5/3 = <u>3.5</u>

(please mark brainliest if this helps!)

8 0
3 years ago
The solution for x2+2x+8&lt;0
Tema [17]
Answer for this question is the zeros are greater than -4 and 2
7 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
A $50 item was marked up 20%. What is the total increased cost of the item?​
GenaCL600 [577]

They increased it by $10.

It ended up costing $60.

Hope this helps.

4 0
3 years ago
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