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Andrew [12]
3 years ago
7

Please help

Mathematics
2 answers:
lys-0071 [83]3 years ago
8 0

Answer:

D.Negative 12 squared +11st minus 2t squared

Step-by-step explanation:

We are given that

(-3s+2t)(4s-t)

We have to find the value of product.

(-3x+2t)(4s-t)

Multiply each term of first bracket with second bracket

-3s(4s-t)+2t(4s-t)

Now, after multiplying we get

-12s^2+3st+8st-2t^2

Combine like terms then, we get

-12s^2+11st-2t^2

Hence,the product

(-3x+2t)(4s-t)=-12s^2+11st-2t^2

Option D is true.

maxonik [38]3 years ago
7 0

Answer:

D. -12s squared + 11st - 2t squared

Step-by-step explanation:

FOIL:

-3s(4s) - 3s(-t) + 2t(4s) + 2t(-t)

- 12s² + 3st + 8st - 2t²

Simplify.

- 12s² + 11st - 2t²

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Ivenika [448]
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When x increases by 1, y decreases by 2, so the slope is -2.

The equation of the function is
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BRAINLIEST !!! 14. Find the length of the midsegment (NOT X)<br> 88+40<br> 16x<br> 7
VARVARA [1.3K]

Answer:

Step-by-step explanation:

The length of the midsegment is  half of the base in this case.

12x = 8x + 40

4x=40

x=10

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7 0
3 years ago
Read 2 more answers
5(d+2)=15 (will mark brainliest)​
Aleonysh [2.5K]

Answer:

d = 1

Step-by-step explanation:

5(d+2)=15 (Given)

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5d = 5 (subtraction)

d = 1 (division)

3 0
3 years ago
An ordinary fair die is a cube with the numbers one through six on the sides represented by painted that. Imagine that such a di
Solnce55 [7]

Answer:

ok so what i think your trying to ask is if we roll two dice that the sum will be more then 6

Two dice

Assuming that the dice are unbiased or not " loaded".

Each side has the same probability, is 1/6 =0.16667, to turn up when rolled, if the die (D) is unbiased. The probability of a side turning up on D1 when 2 dice ( D1,D2) are rolled, is independent of the side turning up in D2. So this is an independent event.

How many ways can one get a sum total of 6 if D1 &D2 are rolled at the same time?

These are the possibilities

Case 1.

D1 =1 & D2=5

Or

D1= 5 & D2=1

Case 2.

D1 =2 & D2=4

Or

D1= 4 & D2= 2

Case 3. D1=3, D2=3

P3 =0.027778

Let's say, P 1 the probability for case 1 and P2 for case 2. There are no other cases.

The final probability P and is the sum total P = P1 + P2 + P3 the probability law of mutually exclusive events.

P1= 0.02778+ 0.02778 =0.055558

P2= 0.02778+0.02778 =0.055558

Same way,

P3=0.027778, when there is only one way to get the sum 6.

So, P = 0.138894

Based on truncating at the sixth decimal place.

A visual representation with two unbiased dice and the possible cases would also give the same result and is a short cut method. I like to derive from the basics.

Hope This Helps!!!

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Answer: 805 tens would be 8050

Step-by-step explanation:

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