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ryzh [129]
3 years ago
6

Solve -2(4y + 6) = 52. 5 -5 8 -8 please help and hurry!!!!!!!!!!!

Mathematics
1 answer:
scoray [572]3 years ago
3 0
-2(4y+6) = 52
-8y -12 = 52
Add 12 on both sides

-8y= 64
Y= -8

Hope this helps!
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Neil took his family of five out to dinner at HogTastic Pit BBQ. They all ordered the buffet and drink at $12.80 each. Neil left
aliina [53]
$12.80 × 1.075 = $13.76

$13.76 × 1.15 =$15.82
5 0
3 years ago
Zack has two strings of equal length. One string is red the other is yellow. After cutting 2.5 meters of the red string and 3.8
ddd [48]

Answer:

6.4 m

Step-by-step explanation:

We have 2 expressions here.  The first one is the fact that r = y.  That's one of 2 equations.  The second one involves whats' left after cutting off certain lengths of each color string.  We cut 2.5 m from red, we cut 3.8 m from yellow.  We know that what's left of red is 1.5 times the length of what's left of yellow.  What's left of red is r - 2.5; what's left of yellow is y - 3.8.  We know that r = 1.5y, so filling that in with our corresponding expressions gives us

r - 2.5 = 1.5(y - 3.8)

Distribute to get

r - 25 = 1.5y - 3.2

Now from the first expression, r = y, so fill in y for r to get an equation in one variable:

y - 2.5 = 1.5y - 3.2

Combine like terms:

-.5y = -3.2 and divide to get

y = 6.4

Check it to make sure it works.  What's left of red should be 1.5 times the length of what's left of yellow and y = 6.4:

What's left of red: 6.4 - 2.5 = 3.9

What's left of yellow: 6.4 - 3.8 = 2.6

1.5 x 2.6 = 3.9, just like it should!

4 0
3 years ago
What is the 20th term of 64,67,70,73,76
PIT_PIT [208]

Let's analyze the increments for each step of the sequence:

Each step we add 3 to the previous number.

Since we need the 20th, from what we saw, in the 20th term we will have added three 20 times.

So the procedure is: calculate how much is 3 times 20 and then add that to the first term of the sequence.

3\times20=60

So we add 60 to the first term to find the 20th term:

64+60=124

Answer: 124

6 0
1 year ago
Somebody please assist me here
Anettt [7]

The base case of n=1 is trivially true, since

\displaystyle P\left(\bigcup_{i=1}^1 E_i\right) = P(E_1) = \sum_{i=1}^1 P(E_i)

but I think the case of n=2 may be a bit more convincing in this role. We have by the inclusion/exclusion principle

\displaystyle P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1 \cup E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le P(E_1) + P(E_2) \\\\ P\left(\bigcup_{i=1}^2 E_i\right) \le \sum_{i=1}^2 P(E_i)

with equality if E_1\cap E_2=\emptyset.

Now assume the case of n=k is true, that

\displaystyle P\left(\bigcup_{i=1}^k E_i\right) \le \sum_{i=1}^k P(E_i)

We want to use this to prove the claim for n=k+1, that

\displaystyle P\left(\bigcup_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

The I/EP tells us

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cup E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right)

and by the same argument as in the n=2 case, this leads to

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) = P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) - P\left(\left(\bigcup\limits_{i=1}^k E_i\right) \cap E_{k+1}\right) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1})

By the induction hypothesis, we have an upper bound for the probability of the union of the E_1 through E_k. The result follows.

\displaystyle P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le P\left(\bigcup\limits_{i=1}^k E_i\right) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^k P(E_i) + P(E_{k+1}) \\\\ P\left(\bigcup\limits_{i=1}^{k+1} E_i\right) \le \sum_{i=1}^{k+1} P(E_i)

5 0
2 years ago
X square + 3 x minus 9 ÷× square - 9 X + 3 ​
olganol [36]

Answer:

1

Step-by-step explanation:

8 0
3 years ago
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