In 1980 Kim was 16 and Anna was 8. When was Kim age four times Anna's age?

Mathematics

1 answer:

max2010maxim [7]3 years ago

3 0

Answer:

48 and prob in 1984

Step-by-step explanation:

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How do you do this question?

LiRa [457]

Step-by-step explanation:

arctan(x) is less than π/2 for x > 0. Therefore, if we say g(x) = 13 (π/2) / eˣ, then f(x) < g(x) for all values of x > 0.

g(x) = 13 (π/2) / eˣ

g(x) = 13π/2 · (1/e)ˣ

So g(x) is a geometric series where r = 1/e. Since |r| < 1, the series converges.

Since g(x) converges, the smaller f(x) also converges.

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4 years ago

A) (-3, 0)<br> B) (0, 3)<br> C) (0, -3)<br> Please help

Tom [10]

Why is the point (4, 3) there but not mentioned??
Anyways, the point that P (3, 0) is after rotating 180° about the origin is (-3, 0).
btw, for 180 rotations about O, just take negative value of (x, y)--> (-x, -y)
That way -(3), -(0) becomes (-3, 0)

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3 years ago

A deep sea diver descended from a rock that is 45 feet below sea level to a coral reef that is 88 feet below sea level.How far d

Veseljchak [2.6K]

Just take the difference of 88 and 45.
43 feet

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3 years ago

Read 2 more answers

What is the area of the logo?

Ludmilka [50]

Answer:

$the \: area \: of \: the \: logo \: is \: {91cm}^{2}$

Step-by-step explanation:

To determine the area of the logo you have to calculate the area of the triangle and the square that comform it and then add the four areas.

Area of the square.

To calculate the area of the square you have to calculate the square of one of its sides, following the formula:

$a = {a}^{2}$

Where "a" is the length of one of its side.

The side length of the square is a=7cm, so its area will be:

$asquare = {7}^{2} \\ asquare = {49cm}^{2}$

Area of the triangles.

The three triangles are equal, they have a base equal to the side of the square, i.e. with a length of 7cm, and their height is h=4cm.

To calculate the area of one triangle, you have to multiply its base by its height and divide by 2, following the formula:

$a = \frac{b.h}{2} \\ a = \frac{7.4}{2} \\ a = \frac{28}{2} \\ a = {14cm}^{2}$

The area calculated the correspond to one triangle, since all triangles are congruent, you have to multiply the said area by 3 to determine the area of three figures:

$atriangles = 3a \\ atriangles = {42cm}^{2}$