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miskamm [114]
3 years ago
5

How to tell if a equation has one solution, no solution, or an infinite number of solutions

Mathematics
1 answer:
Vadim26 [7]3 years ago
3 0

Answer:

no solution

Step-by-step explanation:

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Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

8 0
3 years ago
20,157.5 + 8,725.37 =
Anna [14]

Answer:

28,882.87

Step-by-step explanation:

 20,157.50

+   8,725.37

  28,882.87

5 0
3 years ago
Read 2 more answers
How many milligrams are in 1/2 tsp
lutik1710 [3]
5,687.50 mg are in 1/2 a tsp
5 0
3 years ago
Read 2 more answers
When I double my number, add three, subtract my number, and subtract one, I get my number plus two.
Umnica [9.8K]

Answer:1=1

Step-by-step explanation:Let your no.= x

double the number=2x

add 4 = 2x+4

subtract same number=2x+4-x

subtract 3=2x+4-x-3

I get my number plus 1=x+1

so equation we get

2x+4-x-3=x+1

x+1=x+1

x=x

it means  x is equal to all real number

So your number can be any real number

8 0
3 years ago
The sum of a number and twice another number is 17. The sum of twice the first number and the second number is 4. What are the t
zhannawk [14.2K]

Step-by-step explanation:

Given that,

a + 2b = 17

Also,

2a + b = 4

By adding the above expressions we get,

a + 2b + 2a + b = 17 + 4

3a + 3b = 21.

3 (a + b) = 21

a + b = 21/3

a + b = 7 Let us take 'b' = 10

a + 10 = 7

a = 7 - 10

a = -3

Now let's check by verifying by putting a = -3 and b = 10 in the previous expressions

a + 2b = 17

-3 + (2 × 10) = 17

-3 + 20 = 17

17 = 17

2a + b = 4

(2 × -3) + 10 = 4

-6 + 10 = 4

4 = 4

Therefore, a = -3 and b = 10 [(-3,10)] is the solution.

8 0
3 years ago
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