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lord [1]
3 years ago
7

Solve the quadratic equation by factoring. 9x^2+24x+20=4

Mathematics
1 answer:
const2013 [10]3 years ago
5 0

D. -4/3

Let's solve your equation step-by-step.

9x2+24x+20=4

Step 1: Subtract 4 from both sides.

9x2+24x+20−4=4−4

9x2+24x+16=0

Step 2: Factor left side of equation.

(3x+4)(3x+4)=0

Step 3: Set factors equal to 0.

3x+4=0 or 3x+4=0

x= −4 /3

Hope This helps

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A circular mirror has a diameter of 12 inches. part a what is the area, in square inches, of the mirror?
vovangra [49]
1st part is c
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6 0
3 years ago
A frog eats 8 flies in 4 minutes. How many flies per minute does the frog eat
Liono4ka [1.6K]

Answer:

2 per minute

Step-by-step explanation:

Because 8 divided by 4 is 2.

3 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
What is the missing length?<br> 3 yd<br> 6 yd<br> area = 39 yd 2
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Answer:

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Step-by-step explanation:

3 0
3 years ago
For a club gathering you were having, you bought a total of 50 burgers, and spent $90. You paid $2 per turkey burger, and $1.50
Sedaia [141]

Answer:

<u></u>

  • <u>No. You would have to cut the number of veggie burgers in more than half.</u>

Explanation:

<u>1. Model the situation with a system of equations</u>

<u />

<u>a) Name the variables:</u>

  • number of turkey burgers: t
  • number of veggie burgers: v

<u />

<u>b) Number of burgers:</u>

  • 50 = t + v

<u />

<u>c) Cost of the 50 burgers:</u>

  • $90 = 2t + 1.50v

<u>2. Solve that system of equations:</u>

<u />

<u>a) System</u>

  • 50 = t + v
  • 90 = 2t + 1.50v

<u>b) Mutliply the first equation by 2 and subtract the second equation</u>

  • 100 = 2t + 2v
  •  90 = 2t + 1.50v

  • 10 = 0.5v
  • v = 20 ⇒ t = 50 - 20 = 30

<u />

<u>c) How much would you spend if the next year you buy the double of 20 turkey burgers (40) and the half of 30 veggie burgers (15)</u>

  • $2(40) + $1.50(15) = $80 + $22.50 = $102.50

Then, you if you double the number of turkey burgers, and cut the number burgers in half, you would spend more than $90 ($102.50).

6 0
4 years ago
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