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2 years ago

Write down the first even multiple of 7

Mathematics

1 answer:

Fiesta28 [93]2 years ago

6 0

Answer:

Step-by-step explanation:

multiples are the list of number that can be multiplied by a certain number.

list the multiples

make a chart

1 × 7 = 7 (odd)

2 × 7 = 14 (even)

the first even multiple is 14 (2×7)

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4. Seven times the difference of a number k and five is twenty-onel

laila [671]

Answer:

k = 8

Step-by-step explanation:

7(k - 5) = 21

k - 5 = 3

k = 8

Hope this helps!

4 0

3 years ago

What is the fourth term of the arithmetic sequence described by the explicit formula an = 2 + (n - 1)25?

Elis [28]

Answer:

Step-by-step explanation:

plug in 3 if we're going by 0,1,2,3

2+ (3-1)25

2+50

3 0

3 years ago

Help fast!! please.

alexira [117]

Answer:

Mean = 10

Variance = 30.89

Standard Deviation = 5.56

Step-by-step explanation:

Find the Mean, Variance and Standard deviation

6, 14, 8, 2, 11, 7, 12, 4, 18, 18

Step 1

We rearrange the data and find the Mean

2, 4, 6, 7, 8, 11, 12, 14, 18, 18

Mean = Sum of values/Number of values

= 2 + 4+ 6+ 7+ 8 + 11 + 12 + 14 + 18 +18/10

= 100/10

= 10

Step 2

We find the variance of this sample days ma

Variance = (x - Mean)²/n - 1

n = 10

Hence,

=( (2 - 10)² +(4 - 10)²+ (6 - 10)²+ (7 - 10)² + (8 - 10)² + (11 - 10)² + (12 - 10)² + (14 - 10)² + (18 - 10)² +(18 - 10)²) /10 -1

= ( 64 + 36 + 16 + 9 + 4 + 1 + 4 + 16 + 64 +64)/9

= 278/9

= 30.88888889

Approximately = 30.89

Step 3

Standard Deviation = √Variance

= √30.89

= 5.5578772926

Approximately = 5.56

6 0

3 years ago

P(x)=Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,12).

Mila [183]

Answer:

The polynomial is:

$p(x) = -x^3 - 2x^2 + 5x + 6$

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$ , in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that $x_1 = -3, x_2 = -1, x_3 = 2$ . Thus

$p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)$

$p(x) = a(x - (-3))*(x - (-1))*(x-2)$

$p(x) = a(x+3)(x+1)(x-2)$

$p(x) = a(x^2+4x+3)(x-2)$

$p(x) = a(x^3+2x^2-5x-6)$

Passes through the point (1,12).

This means that when $x = 1, p(x) = 12$ . We use this to find a.

$12 = a(1 + 2 - 5 - 6)$

$-12a = 12$

$a = -\frac{12}{12}$

$a = -1$

Thus

$p(x) = -(x^3+2x^2-5x-6)$

$p(x) = -x^3 - 2x^2 + 5x + 6$

6 0

3 years ago

Can anyone please help me?

sasho [114]

Answer: I can't see if

Step-by-step explanation: Hope this notifys u

6 0

3 years ago

Read 2 more answers