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mafiozo [28]
3 years ago
11

What is 20 percent of 70 plants

Mathematics
2 answers:
g100num [7]3 years ago
8 0
To find the percent of a number, multiply seventy by the decimal for of the percent.

Remember, 20% = .20

.2 • 70 = 14

ANSWER: 14 plants

BONUS: If you are given two numbers, say, "What percent of 70 is 14?"
Simply divide 14 by the original number, 70.
14/70 = .2 = 20%

This is also a good way to check your answers.

Hope this helps you! (:
Greeley [361]3 years ago
7 0
18 plants i think is the answer

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Graph the function f(x) = 3x-5
nexus9112 [7]
Have a nice day :)

f(0)=3*0-5=-5
3x-5=0
3x=5
x=5/3

=====================================================

5 0
2 years ago
How do you factor 3x^2-7x-2
vivado [14]

3x^2-7x-2

doesn't factor


You can not factor this


I hope that's help !

3 0
3 years ago
Circle A has a diameter of 9 inches, a circumference of 28.26 inches, and an area of 63.585 square inches. The diameter of circl
sergiy2304 [10]
A) The equation for circumference is C=2piR. So Filling in for circle A we have 28.26=2*pi*4.5 so we want to isolate pi which I'm gonna call x for it's easier for me xD. So we're gonna start by dividing 4.5 from each side which is gonna leave us with 6.28=2*x which gives you x(pi)= 3.14. For circle B we have 15.70=2*x*2.5 isolate x by first dividing 2.5 which leaves us again with 6.28=2x and x= 3.14. 
B) The equation for area is A=piR^2. So again for circle A we have 63.585=x9^2. This one is harder but also are you sure that the area is 63.585 it's supposed to be 254.469 (We'll come back to this)
C) The observation you can make about the value of pi for circles A and B is that it stays consistent at 3.14
8 0
3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
What is 21/11 as a mixed fraction
castortr0y [4]

Answer:

the answer is

1 10/11

4 0
3 years ago
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