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2 years ago

13

The selling price of a shirt before sales tax 10$.the finial price of a shirt including sales tax is 12.5 .what is the rate of t

he sales tax applied

1 answer:

maw [93]2 years ago

7 0

$11.25 because 10 times 1.125 is equal to 11.25

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Hugo is mixing blue paint with red paint to create purple paint. The ratio of blue to red is 3 to 2. How many pints of blue pain

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From my calculations I got :
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- 16 pints of red paint

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2 years ago

A soccer team played 160 games and won 65 percent of them. How many games did it win ? Please show work

galben [10]

To find the percent of games the soccer team won, you would do 160 multiplied by 0.65, because if you were to convert the 65% into a decimal, you would get 0.65. Notice the key word, "of". This refers to multiplication. So when doing 160*0.65, you get 104 games won. Therefore the soccer team won 104 of their games.

Hope this helps! :D Feel free to ask more questions, or ask me questions about my explanation.

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Which postulate or theorem, if any, could be used to prove the triangles congruent? If not

inysia [295]

Question 11a)

We are given side BC equals to side CE and angle CBA equals to angle CED
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We have enough information to deduce that triangle ABC and triangle CDE are equal by postulate Angle-Side-Angle (ASA)

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Question 11b)

We are given side AB equal to side ED, side BC equals to side EF, and side AC equals to side DF
We have enough information to deduce that triangle ABC and triangle DEF congruent by postulate Side-Side-Side (SSS)

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Question 11c)

We are given side AC equals to side DF, angle ABC equals to angle DEF, and angle BAC equals to angle EDF

We have enough information to deduce that triangle ABC congruent to triangle DEF by postulate Angle-Side-Angle (ASA)

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Question 11d)

We do not have enough information to tell whether this shape congruent or not

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2 years ago

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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 $-4a^3b$ .

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2 years ago