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ra1l [238]
3 years ago
8

In standard form I don’t know exactly

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0
X=-3. That’s the answer
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Anthony got a new iPhone. He needs to create a 4-digit passcode. How many passcodes can be created if REPETITION OF DIGITS IS AL
Snezhnost [94]
An infinite amount...since there ARE and infinite amount of number in the universe.

1111

2222

3333

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5555

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7777

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1010

1234

4321

5432

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7 0
3 years ago
What is the product of (x-2)(x+2)?
Ilya [14]

Answer:

The difference of squares is (a - b)(a + b) = a² - b². Since a = x and b = 2, the answer is x² - 4.

6 0
3 years ago
Read 2 more answers
What value of x satisfies the equation below? Explain<br> how you know<br> -3x = 15
lisov135 [29]
-3x = 15
————
-3. -3

x= -5
5 0
3 years ago
Read 2 more answers
A coin will be tossed three times, and each toss will be recorded as heads (H) or tails (T). Give the sample space describing al
cricket20 [7]

Answer:

Sample space:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Third toss is heads:

{HHT, HTT, THT, TTT}

First toss is heads, second tails and third heads:

{HTH}

Explanation:

In order to find the sample space we need to write all the possible outcomes from tossing the coin three times.You can start by writting all the possible outcomes for the first toss being a head and finish with all the possible outcomes for the first toss being Tails. The number of elements in the sample space can be found by multiplying the possible outcomes of each toss together, so 2*2*2=8

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

For the third toss being heads we will have one restriction this time. Which is that the third toss is Tails. You can take the elements from the sample space where the third element is a tails. In this case, the number of elements can be calculated by multiplying the possible outcomes for only the first two tosses 2*2=4

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

For first toss is heads, second tails and third heads there is only one possible way for you to get this outcome so there is only one element to this set.

{HTH}

4 0
2 years ago
L=U²A divided by (A+B)² make A the subject of formular​
nikklg [1K]

Answer:

A =  [- 2Bl + U^2 +/- √((2BL - U^2)^2 - 4B^2L^2) ] / 2L

Step-by-step explanation:

L = U^2A  / (A + B)^2

U^2 A = L(A^2 + 2AB + B^2)

U^2 A = LA^2 + 2ABL + B^2L

LA^2 + 2ABL - U^2A + B^2 L = 0

LA^2 + (2BL - U^2)A + B^2L = 0

Using the quadratic formula

A =  [- (2BL - U^2) +/- √[(2BL - U^2)^2 - 4*L*B^2L] / 2L

A =  [- 2BL + U^2 +/- √[(2BL - U^2)^2 - 4B^2L^2) ] / 2L

6 0
3 years ago
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