MATHEMATICS HIGH SCHOOL

Suppose an investment of $2,300 doubles in value every decade. The function gives the value of the investment after x decades. f(x)=2300*2^x How much is the investment worth after 2 decades?a.$2,355,200
b. $92,000
c. $46,000
d. $9,200

Answers

Answer 1
Answer: f ( x ) = 2,300 * 2^x
After 2 decades:  x = 2.
f ( 2 ) = 2,300 * 2 ^2 = 2,300 * 4 = 9,200
Answer:
D ) $9,200 
Answer 2
Answer:

Answer:

1. no

2. 2,684,354,560 crickets

3. $9,200

4. b

5. 7.5

6.b

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a rectangular field with a area of 8000m2 is enclosed by 400m of fencing. determine dimensions to nearest tenth of a metre

Answers

field:x\ *\ y\n\n \left\{\begin{array}{ccc}x\cdot y=8000\n2x+2y=400&/:2\end{array}\right\n\left\{\begin{array}{ccc}x\cdot y=8000\nx+y=200&\to x=200-y\end{array}\right\n\nsubstitute\ to\ x\cdot y=8000\n\n(200-y)\cdot y=8000\n\n-y^2+200y-8000=0\n\n\Delta=b^2-4ac;\ \Delta=200^2-4\cdot(-1)\cdot(-8000)=40000-32000=8000\n\ny_1=(-b-\sqrt\Delta)/(2a);\ y_2=(-b+\sqrt\Delta)/(2a)

\sqrt\Delta=√(8000)=√(1600\cdot5)=40\sqrt5\n\ny_1=(-200-40\sqrt5)/(2\cdot(-1))=(-200-40\sqrt5)/(-2)=100+20\sqrt5\approx144.7\ (m)\n\ny_2=(-200+40\sqrt5)/(2\cdot(-1))=(-200+40\sqrt5)/(-2)=100-20\sqrt5\approx55.3\ (m)\n\nx_1\approx200-144.7=55.3\ (m)\n\nx_2\approx200-55.3=144.7\ (m)\n\nAnswer:Dimension of the field is:55.3m\ *\ 144.7m

I Know this is an extreemlly easy question but i am ssssoo tierd please help WILL REPORT AND GIVE BRAINLYEST1/6 into a decimal and percent

Answers

Answer:

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What is the equation y=2-3x in standard form

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