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Kamila [148]
3 years ago
8

I need help with the top one

Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0
420 remaining = 15%
420 / 0.15 = 2800

answer
2800 trees 
You might be interested in
3. Express the mixed recurring decimal 15.732 in p/q form.​
Alborosie

Answer

15717/999

Step-by-step explanation

n=15.732732

10n=157.32732

100n=1573.27327

1000n=15732.732732

n= 15.732732

999n=15717

999 999

4 0
2 years ago
Read 2 more answers
11
LuckyWell [14K]

Answer:

Not sure sorry Im really bad at this stuff

Step-by-step explanation:

8 0
3 years ago
-6x-16 < 2x please help asap! and help me solve it
ICE Princess25 [194]

1. Divide both sides by 2

2. Add 16 to each side

Answer  x > -2

         

3 0
3 years ago
he isosceles triangle has a perimeter of 7.5 m. Which equation can be used to find the value of x if the shortest side, y, measu
lapo4ka [179]

Answer:

2x+ 2.1 = 7.5

x=2.7

Step-by-step explanation:

The perimeter of an isosceles triangle with sides x and y is

2x+y = P

I put y as the single side since the question said y is the shortest side.  It should say shorter sides if the two sides with the same length are smaller than the third.

We know y =2.1 and P = 7.5

Substituting these values in, we get

2x+ 2.1 = 7.5

Subtract 2.1 from each side

2x+2.1 -2.1 = 7.5 -2.1

2x= 5.4

Divide each side by 2

2x/2 = 5.4/2

x = 2.7 m



3 0
3 years ago
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:

<u>The Derivative of a Function</u>

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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