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3 years ago

6

using the standard form ax^2+bx+c=0 of the given quadratic equation, factor the left hand side of the equation into two linear f

actors.

1 answer:

Galina-37 [17]3 years ago

5 0

Answer:

See below

Step-by-step explanation:

Since you didn't give an actual quadratic equation and just the standard form equation, I will reduce ax^2+bx+c=0:

(ax+p)(x+q) where p+q=b and pq=c

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Find the mean of this data set: {– 78, -12, 3, – 44}

blagie [28]

Answer:

<u>Answer</u><u>:</u><u> </u><u>A</u><u>.</u><u> </u><u>-</u><u>3</u><u>2</u><u>.</u><u>7</u><u>5</u>

$\dashrightarrow{ \tt{mean = \frac{ \sum f(n)}{n} }} \\$

• f(n) are the values of the set

• n is the number of items

$\dashrightarrow \: { \tt{mean = \frac{ - 78 + ( - 12) + 3 + ( - 44)}{4} }} \\ \\ \dashrightarrow \: { \tt{ \frac{ - 131}{4} }} \\ \\ \dashrightarrow \: { \boxed{ \tt{mean = - 32.75}}}$

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3 years ago

In two or more complete sentences, compare the number of x-intercepts in the graph of f(t) = t2 to the number of x-intercepts in

Strike441 [17]

F rests at (0/0) and grows upwards, so there is only a single x-intercept. g is f moved 3 to the left, so it also only has one intercept but at (-3,0).

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3 years ago

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suzy uses 1/2 cup of raisins for every 1 1/4 cup of sugar cookies. How many sugar cookies does Suzy need for 2 cups of raisins?

Mandarinka [93]

5 cups.

To go from 1/2 to 2 cups of raisins you multiply by 4.
So 1 1/4 is really 5/4. That times 4 is 20/4 which is 5.

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3 years ago

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The sum of three decimals.<br> 23.56 2.56 2.5892.<br><br> Step by step I need to show my work.

Volgvan

Answer:

28.7092

Step-by-step explanation:

2.5892 + 23.5600 = 26.1492

26.1492 = 28.7092

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A survey was conducted of 42 students who take math, Eng lish, or history. In the survey, 15 students said they take math and hi

strojnjashka [21]

Answer:

16 students take only English or only Math.

Step-by-step explanation:

We can solve this problem by treating these values as sets, and building the Venn Diagram.

I am going to say that:

A is the number of students who take Math.

B is the number of students who take English.

C is the number of students who take History.

We have that:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a is the number of students that only take Math, $A \cap B$ is the number of students who take both Math and English, $A \cap C$ is the number of students that take both Math and History, and $A \cap B \cap C$ is the number of students that take all these classes.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following subsets:

$a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)$

There were 42 students suveyed. This means that:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 42$

We start finding the values from the intersection of three sets.

12 students said they take math, English, and history. This means that:

$A \cap B \cap C = 12$

18 students said they take English and history. This also takes into account those who take math, english and history. So:

$(B \cap C) + (A \cap B \cap C) = 18$

$B \cap C = 6$

17 students said they take math and English.

$(A \cap B) + (A \cap B \cap C) = 17$

$A \cap B = 5$

15 students said they take math and history

$(A \cap C) + (A \cap B \cap C) = 15$

$A \cap C = 3$

2 students said they only take history.

$c = 2$

How many students take only English or only math?

This is a + b, that we can find by the following formula:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 42$

$a + b + 2 + 5 + 3 + 6 + 12 = 42$

$a + b = 16$

16 students take only English or only Math.

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3 years ago