3 years ago

(i) Which of the following is an empty

Mathematics

1 answer:

lakkis [162]3 years ago

6 0

Answer:

Step-by-step explanation:

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Simplify 3(x-6)-4(x+3)

Finger [1]

Answer:

Step-by-step explanation:

look this up it give you the answer

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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).

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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

(Less than)

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