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Bess [88]
3 years ago
13

Multiply. −2x(6 x 4 −7 x 2 +x−5) Express the answer in standard form.

Mathematics
1 answer:
Flura [38]3 years ago
3 0

Answer:

The answer is  -22x*2+10x

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Please help:<br> In PQR, sin P=64, sin R=35, and r=21. Find the length of p.
son4ous [18]

Answer:

32.9 units

Step-by-step explanation:

p/sinP = r/sinR

p = sinP × r/sinR

p = sin(64) × 21/sin(35)

p = 32.9069916

5 0
3 years ago
654x67<br> Quien me ayuda please
Likurg_2 [28]
The answer is 43,818
3 0
3 years ago
Read 2 more answers
Simplify the expression.<br> 7x2 + 3 - 5(x2 - 4)<br> 2x2-17<br> 2x2-1<br> 2x2 + 23<br> 25x2
Blizzard [7]

Answer:

1. 14+3-5(-2)= 17+10=27

2. 14-17= -3

3. 4-1=3

4. 4+23=27

5. 50

3 0
3 years ago
. Quantas senhas com 4 algarismos diferentes podemos escrever com os algarismos 1, 2, 3, 4, 5, 6?
ruslelena [56]

Answer:

We can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

Step-by-step explanation:

We have to find how many passwords with 4 different digits can we write with the numbers 1, 2, 3, 4, 5, and 6.

Firstly, it must be known here that to calculate the above situation we have to use Permutation and not combination because here the order of the numbers in a password matter.

Since we are given six numbers (1, 2, 3, 4, 5, and 6) and have to make 4 different digits passwords.

  • Now, for first digit of the password, we have 6 possibilities (numbers from 1 to 6).
  • Similarly, for second digit of the password, we have 5 possibilities (because one number from 1 to 6 has been used above and it can't be repeated).
  • Similarly, for the third digit of the password, we have 4 possibilities (because two numbers from 1 to 6 have been used above and they can't be repeated).
  • Similarly, for the fourth digit of the password, we have 3 possibilities (because three numbers from 1 to 6 have been used above and they can't be repeated).

So, the number of passwords with 4 different digits we can write = 6 \times 5 \times 4 \times 3  = 360 possibilities.

Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

4 0
2 years ago
Among 35 students in a class, 20 of them earned A's on the midterm exam, 13 earned A's on the nal exam, and 12 did not earn A's
enyata [817]

Answer:

The correct answer is "\frac{2}{7}".

Step-by-step explanation:

According to the question,

Number of students,

= 35

A ! mid term A,

n(A) = 20

B : final A,

n(B) = 13

Didn't start,

n(A^c \cap B^c) = 12

Now,

⇒ n(A \cup B)=35-12

                   =23

then,

⇒ n(A \cup B)=20+13-23

                   =10

hence,

The probability will be:

⇒ P(A \cap B)=\frac{10}{35}

⇒                 =\frac{2}{7}

7 0
3 years ago
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