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3 years ago

10

A rectangle in quadrant I rotates 270° counterclockwise. In what quadrant will the transformation lie I rotates 270° countercloc

kwise in which quadrant will the transformation lie?

1 answer:

Alex Ar [27]3 years ago

5 0

Quadrant IV.

Every 90° counterclockwise moves it one quadrant further. 270÷90=3, 1+3=4, therefore the transformation will be in quadrant IV

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I need help how to solve it mot the answer plz and thank u

jek_recluse [69]

The second one would be the answer. You want to plug in for this question. You can divide over the five leaving you with (x-2)^2 =25. Since you are squaring this answer, you want to get either 5 or negative five. Both -3 and 7 will do that.

4 0

3 years ago

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valentina_108 [34]

Y= 1/2x+7

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3 0

3 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

3 years ago

Could 18 in, 6 in, 13 in form a triangle

Tanya [424]

Answer: Yes

Step-by-step explanation:

The reason on why it could form a triangle because 18 is the highest side in the triangle and if you have 2 of the smaller sides that are greater than the larger size then you could form a triangle and 13 and 6 equals to 19 so you could make a triangle hopes this helps and have a good day.

7 0

3 years ago

What is the radius and diameter of the circle 14

gizmo_the_mogwai [7]

Answer:

it is

D=14

R=7

Step-by-step explanation:

sorry if it is incorrect

3 0

4 years ago