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2 years ago

OK PLEASE HELP ME GUYS I ALWAYS HELP YOU so here it is

1 answer:

8 0

F=P(1/2)^(t/h)
F=future amount
P=present amount
t=time elapsed
h=legnth of half life

P=96
t=2
h=1
F=96(1/2)^(2/1)
F=96(1/2)^2
F=96(1/4)
F=96/4
F=24 grams

grow by 35%
compound interest
F=P(1+rate)^time
F=95000(1+0.35)^10
F=95000(1.35)^10
F=95000(20.106555868618)
F=1910122.9075187

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A horizontal line is seven which of the following is true

nydimaria [60]

Answer:What are the answer choices??

Step-by-step explanation:

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3 years ago

Madam Siti bought the same lengths of red and yellow ribbons. After she used 836 cm of the red ribbon, the length of the yellow

Yuliya22 [10]

Answer: 1045 cm of red ribbon and 1045 cm of yellow ribbon

Step-by-step explanation:

Let R represent the amount red ribbon and Y represent the amouont of yellow ribbon.

bought the same lengths of red and yellow: R = Y

after used 286 of red, yellow is 5 times the remaining red; 5(R - 836) = Y

Use substitution method to solve:

5(R - 836) = R

5R - 4180 = R distributed 5 on the left side

4R - 4180 = 0 subtracted R from both sides

4R = 4180 added 4180 to both sides

R = 1045 divided both sides by 4

5 0

3 years ago

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Need help with Calculus 1 inverse trig functions

lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

The Derivative of a Function

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

$\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}$

Where u is a function of x as provided:

$y=3tan^{-1}(x+\sqrt{1+x^2})$

If we set

$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

$\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}$

Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

8 0

3 years ago

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yaroslaw [1]

Answer:

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Step-by-step explanation:

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7 0

2 years ago

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wolverine [178]

Answer:

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Step-by-step explanation:

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6 0

3 years ago

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