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3 years ago

12

in triangle abc , the length of ab is 3 cm and the length of bc is 7 cm. triangle abc will be dilated with the center at the ori

gin to create a’ b’ c’. if image of point a (-3,5) after the dilation is a’ (-12,20) what is the length of b’c’ ?

1 answer:

lys-0071 [83]3 years ago

3 0

Answer:

What are the options?

Step-by-step explanation:

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Help me substitution method define variables

Rina8888 [55]

$\left \{ {{S=14+L} (I) \atop {S+L=124}(II)} \right.$

<span>Substitute (I) into (II), we have:
</span> $S+L=124$
$(14+L)+L=124$
$14 + 2L = 124$
$2L = 124 - 14$
$2L = 110$
$L = \frac{110}{2}$
$\boxed{L = 55}$

<span>Replace (II) in (I), we have:
</span> $S=14+L$
$S = 14+ 55$
$\boxed{S = 69}$

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3 years ago

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3 years ago

A friend wants to buy a pool and has two places she wants to purchase the pool with the largest volume which pool should she buy

slamgirl [31]

Answer:

20'×15 in 54 inches

Step-by-step explanation:

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2 years ago

4 times a number is decreased by 9, the result is the same as when 15 is added to twice the number. find the number

igor_vitrenko [27]

Answer:

12

Step-by-step explanation:

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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 $-4a^3b$ .

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3 years ago