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olga2289 [7]
2 years ago
13

Which of the following equations shows how substitution can be used to solve the following system of equations?

Mathematics
1 answer:
klasskru [66]2 years ago
5 0

Answer:

i have  no idea does anyone else know

Step-by-step explanation:

You might be interested in
What is P(2 or divisor of 9)?<br> please explain!
Dmitrij [34]

The probability that a divisor of 2 or 9 is chosen when the spinner is spun once is 2/3.

<h3>What is the probability?</h3>

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

P(2 or divisor of 9) = section that has a divisor of 2/ total number of sections + section that has a divisor of 2/ total number of sections

1/3 + 1/ 3 = 2/3

To learn more about probability, please check: brainly.com/question/13234031

#SPJ1

3 0
2 years ago
The minimum/maximum value of the function y = a(x − 2)(x − 1) occurs at x = d, what is the value of d?
daser333 [38]

Answer:

d=\frac{3}{2}=1.5

Step-by-step explanation:

We have the function:

y=a(x-2)(x-1)

And we want to find x=d for which the minimum/maximum value will occur.

Notice that our function is a quadratic in factored form.

Remember that the minimum/maximum value always occurs at the vertex point.

And remember that the x-coordinate of the vertex is the axis of symmetry.

Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.

Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.

Our factors are <em>(x-2) </em>and <em>(x-1)</em>.

Therefore, our roots/zeros are <em>x=1, 2</em>.

So, the average of them are:

d=\frac{1+2}{2}=3/2=1.5

Therefore, regardless of the value of <em>a</em>, the minimum/maximum value will occur at <em>x=d=1.5</em>.

Alternative Method:

Of course, we can also expand to confirm our answer. So:

y=a(x^2-2x-x+2)\\y=a(x^2-3x+2)\\y=ax^2-3ax+2a

The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.

And the formula for the vertex is:

x=-\frac{b}{2a}

So, we will substitute <em>-3a</em> for <em>b</em> and <em>a</em> for <em>a</em>. This yields:

x=-\frac{-3a}{2a}=\frac{3}{2}=1.5

Confirming our answer.

4 0
3 years ago
Which equation is a linear function?
Sonja [21]

Answer:

the equation in letter D is a linear function.

6 0
3 years ago
Solve the equation the square of 2x + 1 equals the square of x + 4
gavmur [86]
I am sorry this is where I can get too.

8 0
3 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
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