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2 years ago

Tom swam to the bottom of a swimming pool that was 9 feet deep and he touched the

Mathematics

1 answer:

zmey [24]2 years ago

7 0

Answer:

18 I guess if he can swim straightdown and straight up :-)

Step-by-step explanation:

9+9=18

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AB=45. If RT=3(x+9), what is the value of x?

Margaret [11]

6 because 45 divides by 3 equals 15 and fifteen minus 9 equals 6, if you plug 6 into the equation it makes sense

7 0

3 years ago

Read 2 more answers

Simplify (Are all division) 2+4i 3i 3+2i 4+i 2i^11

Alla [95]

$We\ know:\ i=\sqrt{-1}\to i^2=-1$

$\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i$

$\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i$

$2i^{11}=2i^{10+1}=2i^{10}i^1=2i^{2\cdot5}i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i$

$Used:\ (a+b)(a-b)=a^2-b^2$

6 0

3 years ago

What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?

seraphim [82]

21.6 units

Calculate the length of the 2 inclined sides using the distance formula

d = √(x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 3, 3 ) and (x₂, y₂ ) = (3, 4 )

d = √(3 + 3 )² + (4 - 3)² = √(36 + 1 ) = √37

repeat for (x₁, y₁ ) = (- 3, 3 ) and (x₂, y₂ ) = (3, - 3 )

d = √(3 + 3 )² + (- 3 - 3 )² = √(36 + 36 ) = √72

the vertical side has a length of 7 units

perimeter = √37 + √72 + 7 = 21.6 ( nearest tenth of a unit )

3 0

3 years ago

Please someone help!! I tried my best to complete this problem but i'm stuck now (also am i even doing it right so far??!);(

Alik [6]

Answer and Step-by-step explanation:

You got everything correct so far except for #4.

4. Yes, it is 1. But it would be in months.

So you would put:

1 month = x

12 months = 1 year.

Since the population increases by 1.5 times a month.

For question number 3.

The equation should be:

$f(x) = 100(1.5)^x$ <- Function

$f(x) = 100(1.5)^{12}$ <- Function when x is 12 months (1 year)

(Put those both the same way I put it.)

It gives you the equation to work with, you just have to plug in the values.

1.5 is in the parenthesis because it needs to be the one that is raised by an exponent.

100 is the initial population, so it stays on the outside.

x is the exponent

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%20%3D%20ln%205" id="TexFormula1" title="e^{2x} = ln 5" alt="e^{2x} = ln 5" align=

KiRa [710]

$e ^{2x} = ln5$

Solve for the real domain

$e ^{2x} = ln(5)$

if $f(x) =g(x),$ then $ln(f(x))= ln(g(x))$

$ln(e ^{2x} ) = ln(ln(5))$

Solve : $ln(e ^{2x} ) = ln(ln(5))$

use the logarithmic definition :

$ln(e^{f(x)} ) = f(x)$

$ln(e^{2x} ) = 2x$

$2x=ln(ln(5))$

Divide both sides by 2 :

$\frac{2x}{2} = \frac{ln(ln(5))}{2}$

$x= \frac{ln(ln(5))}{2}$

hope this helps!

8 0

3 years ago