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liraira [26]
2 years ago
9

Can you please help?

Mathematics
1 answer:
docker41 [41]2 years ago
5 0

Answer:

what do you need help with

You might be interested in
What is the maximum number of possible extreme values for the function,<br> f(x) = х2 -7x – 6?
timama [110]

Answer:

Only one extreme value of f(x) is possible.

Step-by-step explanation:

We are given the quadratic function of independent variable x which is f(x) = x² - 7x - 6 ......(1)

Now. the condition for extreme values of f(x) is \frac{df(x)}{dx} =0

Hence, differentiating both sides of equation (1) with respect to x, we get

\frac{df(x)}{dx} = 2x -7 = 0

⇒ x = 3.5.

So there is only one value of x for which f(x) has extreme value which is x = 3.5.

Therefore, only one extreme value of the given function is possible. (Answer)

3 0
3 years ago
2. An expression is shown.
laiz [17]

Adding parentheses in the component 9\div 3 of the expression may bring an output of 48.

<h3>Procedure - Application of hierarchy rules in a arithmetic expression</h3>

In this question we should make use of hierarchy rules represented by the use of parentheses. The parentheses oblige to make operations inside it before making it in the rest of the formula.

Now we decide to add parenthesis in the component 9\div 3 such that the result of the entire expression is 48. We proceed to present the proof:

2 \times 8 \times (9\div 3)

2\times 8 \times 3

2\times 24

48

Adding parentheses in the component 9\div 3 of the expression may bring an output of 48. \blacksquare

<h3>Remarks</h3>

The statement presents mistakes and is poorly formatted. Correct form is shown below:

An expression is shown: 2\times 8 \times 9 \div 3

Using the same expression, add parenthesis so that the value of the expression is 48.

To learn more on hierarchy rule, we kindly invite to check this verified question: brainly.com/question/3572440

5 0
2 years ago
Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability
krok68 [10]

Answer:

P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}

Step-by-step explanation:

Let A and B events. We have defined the probabilities for some events:

P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}

Where A' represent the complement for the event A

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

So for this case we can solve for P(A) like this:

P(A) = 1-P(A') = 1-\frac{68}{73}=\frac{5}{73}

And now we can find P(A \cup B) using the total probability rul given by:

P(A \cup B) = P(A)+P(B) -P(A \cap B)

And if we replace the values given we got:

P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}

And that would be the final answer.

5 0
2 years ago
Given the arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103,...
omeli [17]

We are given

arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103

so, first term is 124

u(1)= 124

now, we can find common difference

d=u(2)-u(1)

d=117-124

d=-7

now, we can find kth term

u(k)=u(1)+(k-1)d

now, we can plug values

and we get

u(k)=124+(k-1)*-7

u(k)=124-7k+7

u(k)=131-7k

u(k) must be negative

so,

u(k)=131-7k

131-7k

now, we can solve for k

7k>131

k>18.714

so, it's closest integer value is

k=19..............Answer


3 0
3 years ago
What lengths of two sides of a triangle are seven and 11. Which could not be the length of the third side?
Finger [1]
You’re answer it would be D.5 because it is the smaller than the other sides
7 0
3 years ago
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