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Anna007 [38]
3 years ago
15

If

x-formula"> and y < 0 ,where is the point (x,y) located?
a) quadrant I
b) quadrant II
c) quadrant III
d) quadrant IV​
Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0

Answer:

Quadrant III

Step-by-step explanation:

x < 0 is the area to the left of the y-axis, and y < 0 is the area beneath the x-axis.  This point is therefore located in Quadrant III.

miss Akunina [59]3 years ago
5 0

Answer:

c) quadrant III

Step-by-step explanation:

I will be quick in this question:

\text {For }y

\text {For }x

What both have in common?

So, the answer is <u>Quadrant III</u><u>.</u>

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I need help !! Can someone answer this mathematics question for algebra 1. Question number five pls.
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Pls help I need this in soon I have less than an hour and will give brainliest to first answer that is correct
lawyer [7]

Answer:

A) Lexi is 15 and Tanner is 18

B) Enrique has 24 magnets and Peter has 88 magnets

Step-by-step explanation:

1.

t = 3 + l

t + l = 33

substitute:

(t) + l = 33

(3 + l) + l = 33

3 + 2l = 33

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solve for t:

t + 15 = 33

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2.

e = p - 64

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4 0
3 years ago
Simplify the following Questions. (IMPORTANT NOTE!!! #'s 2 AND 7 MUST BE IN SCIENTIFIC NOTATION!!!!!) 1.) (x)²(2xy³)^5 2.) (4x10
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Answer:

Step-by-step explanation:

Hint :

(a*b)^{m}= a^{m}b^{m}\\\\\\(a^{m})^{n}=a^{mn}\\\\\\a^{m}*a^{n}=a^{m+n}\\

1)x^{2}(2xy^{3})^{5}=x^{2}*2^{5}*x^{5}*(y^{3})^{5}\\\\\\=x^{2}*2^{5}*x^{5}*y^{3*5}\\\\=x^{2}*2^{5}*x^{5}*y^{15}\\\\=2^{5}*x^{2+5}*y^{15}\\\\=2^{5}x^{7}y^{15}

2) (4x10^{8})^{2}=4^{2}*x^{2}*(10^{8})^{2}\\\\=16*x^{2}*10^{8*2}\\\\=1.6*10*x^{2}*10^{16}\\\\=1.6x^{2}*10^{16+1}\\\\=1.6x^{2}10^{17}

3)(-h^{4})^{5}=(-h)^{4*5}=-h^{20}\\\\\4)(3xy^{3})^{2}(xy)^{6}=3^{2}x^{2}(y^{3})^{2}x^{6}y^{6}\\\\=9x^{2}y^{3*2}x^{6}y^{6}\\\\\\=9x^{2}y^{6}x^{6}y^{6}\\\\=9x^{2+6}y^{6+6}\\\\=9x^{8}y{12}\\\\\\

5) (p^{9})^{-2}=p^{9*-2}=p^{-18}\\\\\\6)(5k^{2})^{3}=5^{3}(k^{2})^{3}=625k^{6}\\\\7)(7x10^{5})^{2}=7^{2}x^{2}(10^{5})^{2}\\\\=49x^{2}10^{5*2}\\\\=4.9*10^{1}x^{2}10^{10}\\\\=4.9x^{2}10^{10+1}\\\\=4.9x^{2}10^{11}

8)x^{3}(-x^{3}y)^{2}=x^{3}*(-x^3})^{2}y^{2}\\\\\\=x^{3}*(-1)^{2}*x^{2*3}y^{2}\\\\=x^{3}*1*x^{6}y^{2}=x^{3+6}y^{2}\\\\=x^{9}y^{2}\\\\\\9)w^{5}(w^{2})^{-4}=w^{5}w^{2*-4}\\\\=w^{5}w^{-8}\\\\=w^{5-8}=w^{-3}\\\\=\frac{1}{w^{3}}\\\\\\10) (2x^{5})^{4}=2^{4}x^{5*4}\\\\=2^{4}x^{20}\\\\=16x^{20}

5 0
3 years ago
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