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frosja888 [35]
3 years ago
12

PLEASE HELP ASAP!!!!!! Will someone please get me ahead and boost my grade up in my math class and my chemistry class? I will do

your history, English, Spanish, or health class in return!!!!!!
Mathematics
2 answers:
Firdavs [7]3 years ago
5 0
What math class are you taking? And what grade are you in currently?
Roman55 [17]3 years ago
4 0
What do you need help on?
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Ok, if you say so. Jean took 1/2 hour to complete 3/8 of a math problem.


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When the function f(x) = 5•2x is evaluated for x = 3, the output is:
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PLEASE HELP!!!!!!!!
vfiekz [6]

Answer:

about 78 years

Step-by-step explanation:

Population

y =ab^t  where a is the initial population and b is 1+the percent of increase  

    t is in years

y = 2000000(1+.04)^t

y = 2000000(1.04)^t

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y = a+bt   where a is the initial population and b is constant increase

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2000000(1.04)^t= 4000000 +500000t

Using graphing technology, (see attached graph  The y axis is in millions of years), where these two lines intersect is the year where food shortages start.

t≈78 years

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3 years ago
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The length of a rectangle is 5 metres less than twice the breadth. If the perimeter is 50 meters,find the length and breadth
MAXImum [283]
<h3><u>S</u><u> </u><u>O</u><u> </u><u>L</u><u> </u><u>U</u><u> </u><u>T</u><u> </u><u>I</u><u> </u><u>O</u><u> </u><u>N</u><u> </u><u>:</u></h3>

As per the given question, it is stated that the length of a rectangle is 5 m less than twice the breadth.

Assumption : Let us assume the length as "l" and width as "b". So,

\twoheadrightarrow \quad\sf{ Length =2(Width)-5}

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

Also, we are given that the perimeter of the rectangle is 50 m. Basically, we need to apply here the formula of perimeter of rectangle which will act as a linear equation here.

\\ \twoheadrightarrow \quad\sf{ Perimeter_{(Rectangle)} = 2(\ell +b) } \\

  • <em>l</em> denotes length
  • <em>b</em> denotes breadth

\\ \twoheadrightarrow \quad\sf{50= 2(2b-5+b)} \\

\\ \twoheadrightarrow \quad\sf{50= 2(3b-5)} \\

\\ \twoheadrightarrow \quad\sf{50= 6b - 10} \\

\\ \twoheadrightarrow \quad\sf{50+10= 6b} \\

\\ \twoheadrightarrow \quad\sf{60= 6b} \\

\\ \twoheadrightarrow \quad\sf{\cancel{\dfrac{60}{6}}=b} \\

\\ \twoheadrightarrow \quad\underline{\bf{10\; m = Width }} \\

Now, finding the length. According to the question,

\twoheadrightarrow \quad\sf{ \ell=(2b-5) \; m}

\twoheadrightarrow \quad\sf{ \ell=2(10)-5\; m}

\twoheadrightarrow \quad\sf{ \ell=20-5\; m}

\\ \twoheadrightarrow \quad\underline{\bf{15\; m = Length }} \\

<u>Therefore</u><u>,</u><u> </u><u>length</u><u> </u><u>and</u><u> </u><u>breadth</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>r</u><u>ectangle</u><u> </u><u>is</u><u> </u><u>1</u><u>5</u><u> </u><u>m</u><u> </u><u>and</u><u> </u><u>10</u><u> </u><u>m</u><u>.</u><u> </u>

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3 years ago
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