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3 years ago

Donna is planning a vacation. She looks at a map with the following scale.

Mathematics

1 answer:

serg [7]3 years ago

8 0

1 inch=1 1/4 mi

13 in = 16 1/4 mi

19 in = 23 3/4 mi

16 1/4 + 23 3/4 =

Answer= 40 mi

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The standard deviation is used in conjunction with the.

Svetlanka [38]

Answer:

it's used in conjunction with variance

4 0

1 year ago

The function c=2.50(n-2)+1.50 represents the cost c in dollars of printing n invitations. Which of the following is not true?

Yuri [45]

Please put the selections. I can't tell you what isn't true if you don't.

6 0

3 years ago

What are the roots of the polynomial equation? x² + 7x + 12 = 0 Enter your answers in the boxes. x1= x2=

oksian1 [2.3K]

The equation factors as
.. (x +3)(x +4) = 0
By the zero-product rule, the roots are
.. x1 = -4
.. x2 = -3

3 0

3 years ago

Read 2 more answers

Gretchen is rolling a number cube. What is the probability of rolling a 1 followed by a 2 or a 3

stira [4]

Answer:

1/18

Step-by-step explanation:

The probability of rolling a 1 is 1/6 because there are 6 numbers. You multiply that probability by the probability of getting a 2 or 3 which is 2/6 or 1/3 to get 1/18.

3 0

2 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago