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babunello [35]
2 years ago
6

What are the x-values of the quadratic equation (x-2)^2-9=0?

Mathematics
1 answer:
skelet666 [1.2K]2 years ago
5 0

Answer: x=−1 or x=5

Step-by-step explanation:

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-3 (v + 4) = 2v - 37
asambeis [7]

Step-by-step explanation:

-3 (v + 4) = 2v - 37

-3v - 12 = 2v - 37

5v = 25

v = 5

3 0
3 years ago
Betty bought two bags full of books at an annual library book sale. If she
almond37 [142]

Answer:

12.90= .90+ 2x

12=2x

6=x

Each bag of books cost was $6

4 0
3 years ago
Read 2 more answers
Prove that the statement (ab)^n=a^n * b^n is true using mathematical induction.
mamaluj [8]

Answer:

see below

Step-by-step explanation:

      (ab)^n=a^n * b^n

We need to show that it is true for n=1

assuming that it is true for n = k;

(ab)^n=a^n * b^n

( ab) ^1 = a^1 * b^1

ab = a * b

ab = ab

Then we need to show that it is true for n = ( k+1)

or (ab)^(k+1)=a^( k+1) * b^( k+1)

Starting with

  (ab)^k=a^k * b^k    given

Multiply each side by ab

ab *  (ab)^k= ab *a^k * b^k

   ( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)

Therefore, the rule is true for every natural number n

5 0
3 years ago
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1+2=<br> 3+4<br> 4=3<br> )+0<br> you dont have to answer just put a random letter or something
Alina [70]

Answer:

3

7

last one doesn't even make sense

3 0
2 years ago
A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5​
gladu [14]

Answer:

A) It will get to a temperature of 125°F at 9:19 PM

B) It will get to a temperature of 150°F at 9:16 PM

C) as time passes temperature approaches the initial temperature of 450°F

Step-by-step explanation:

We are given;

Initial temperature; T_i = 450°F

Room temperature; T_r = 70°F

From Newton's law of cooling, temperature after time (t) is given as;

T(t) = T_r + (T_i - T_r)e^(-kt)

Where k is cooling rate and t is time after the initial temperature.

Now, we are told that After 5​ minutes, the temperature is 300°F.

Thus;

300 = 70 + (450 - 70)e^(-5k)

300 - 70 = 380e^(-5k)

230/380 = e^(-5k)

e^(-5k) = 0.6053

-5k = In 0.6053

-5k = -0.502

k = 0.502/5

k = 0.1004 /min

A) Thus, at temperature of 125°F, we can find the time from;

125 = 70 + (450 - 70)e^(-0.1004t)

125 - 70 = 380e^(-0.1004t)

55/380 = e^(-0.1004t)

In (55/380) = -0.1004t

-0.1004t = -1.9328

t = 1.9328/0.1018

t ≈ 19 minutes

Thus, it will get to a temperature of 125°F at 9:19 PM

B) Thus, at temperature of 150°F, we can find the time from;

150 = 70 + (450 - 70)e^(-0.1004t)

150 - 70 = 380e^(-0.1004t)

80/380 = e^(-0.1004t)

In (80/380) = -0.1004t

-0.1004t = -1.5581

t = 1.5581/0.1004

t ≈ 16 minutes.

Thus, it will get to a temperature of 150°F at 9:16 PM

C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

Thus,we have;

T(t) = T_r + (T_i - T_r)

T_r will cancel out to give;

T(t) = T_i

Thus, as time passes temperature approaches the initial temperature of 450°F

6 0
3 years ago
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