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2 years ago

7y2 - 7y+ 4 -(-3y2 + 7y + 7) (Simplify your answer.)

Mathematics

2 answers:

lesantik [10]2 years ago

7 0

Answer:

10y2 -7y -3

Step-by-step explanation:

7y2 + 3y2= 10y2

-7y

4 - 7=-3

alexandr402 [8]2 years ago

3 0

Answer:

7y + 4
-y - 7

Step-by-step explanation:

1. 7y2 - 7y + 4

7y × 2 = 14y

14y - 7y + 4

= 7y + 4

2. -(-3y2 + 7y + 7)

-3y × 2 = -6y

-(-6y + 7y + 7)

-6y + 7y = 1y = y

= -y - 7

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Answer:

3x(3x+5)(x-2)

Step-by-step explanation:

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3 years ago

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3 years ago

16/__ = -8 ___/12 = -6

Karolina [17]

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3 years ago

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A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a)

marta [7]

(a) If $A,B,C$ are mutually exclusive, then

$P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0$

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

$P(A\cap B)=P(A)P(B),$

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

$P(A\cap B\cap C)=P(A)P(B)P(C)$

so that

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

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

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3 years ago

Of all sixth graders, 70% sent a text message yesterday. Ten trials of a simulation are conducted and the data are recorded belo

Effectus [21]

Answer:

80%

Step-by-step explanation:

According to the given data, the numbers 0 through 6 represent students who sent a text yesterday while the numbers 7 through 9 represent students who did not send a text yesterday.

Based on this key, we will calculate the number of students who sent the text yesterday and the number of students who did not sent the text yesterday.

The given data is: 62072, 34570, 80983, 04292, 83150, 36330, 96268, 14077, 77985, 13511

For each group we will identify how many students sent the text and how many students did not. According to the key, a number from 0 to 6 will be counted as a student who sent the text and a number from 7 to 9 will be counted as a student who did not send the text. Students who did not sent the text are made bold in the groups below.

62072: 4 students sent the text, 1 did not

34570: 4 students sent the text, 1 did not

80983: 2 students sent the text, 3 did not

04292: 4 students sent the text, 1 did not

83150: 4 students sent the text, 1 did not

36330: All 5 students sent the text

96268: 3 students sent the text, 2 did not

14077: 3 students sent the text, 2 did not

77985: 1 student sent the text, 4 did not

13511: All 5 students sent the text

We need to find the probability that 3 or more of a group of 5 students randomly selected will send a text today.

From the given groups, the number of groups where 3 or more students sent a text = 8

This represents our number of favorable or desired outcomes.

Total number of possible outcomes is the total number of groups = 10

Since, probability is defined as the ratio of number of favorable outcomes to total number of outcomes, based on the simulated data the probability that 3 or more of a group of 5 students randomly selected will send a text today will be = $\frac{8}{10}=0.80=80\%$

Thus, based on the simulated data there is 80% probability that 3 or more of a group of 5 students randomly selected will send a text today

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3 years ago

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