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Paha777 [63]
3 years ago
15

(i) Which of the following is an empty

Mathematics
1 answer:
lakkis [162]3 years ago
6 0

Answer:

ok

ok

Step-by-step explanation:

ok

ok

ok

ok

ok

ok

ok

on

ko

k

k

k

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Simplify 3(x-6)-4(x+3)
Finger [1]

Answer:

Step-by-step explanation:

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7 0
3 years ago
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There is a circle with a center of 0,0 on a coordinate plane. There is one point on the circle's circumference in which the x:y
abruzzese [7]

Answer:

Step-by-step explanation:

Suppose the radius is 1. The parametric equations for the circle are

x = cosθ

y = sinθ

x:y = 3:1

tanθ = ⅓

cosθ = 3/√(1²+3²) =  3/√10

sinθ = 1/√10

The solutions are (3/√10, 1/√10) and (-3/√10, -1/√10).

6 0
3 years ago
Is the quotient 3.6÷9 greater or less than one
anastassius [24]
3 6/10÷ 9
36/10×1/9
36/90
.4<1

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Hope this helps!
3 0
3 years ago
Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove
steposvetlana [31]

Answer:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}

Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

8 0
4 years ago
Ubi made a banana cream pie.His brother ate 1/3 of the pie and his sister ate 2/6 of the pie.
tamaranim1 [39]
2/3 of the pie had been eaten, there is 1/3 left
8 0
3 years ago
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