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Mathematics

1 answer:

n200080 [17]2 years ago

3 0

Answer:

both population and sample

Step-by-step explanation:

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Y = ax + 4x - 2 is x= 1. What is the value of a?

puteri [66]

Y = ax + 4x - 2 x= 1
y = a(1) + 4(1) -2
y = a + 2
y - 2 = a

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3 years ago

Solve for x: 3x − 2 > 5x + 10. (5 points)

velikii [3]

Answer:

Option 2.. x > -6

Step-by-step explanation:

3x -5x - 2 > 5x -5x + 10

-2x - 2 + 2 > 10 + 2

-2x/-2 > 12/-2

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2 years ago

Which number is an integer? 1, 0.5, 3/2

aksik [14]

An integer is any whole number (not fractions) that can be either negative or positive.

1 is the only integer because 0.5 and 3/2 are not whole numbers and are fractions, while 1 is a whole number.

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2 years ago

Find a8 of the sequence 10,9.75,9.5,9.25,….

balandron [24]

Answer:

10,9.75,9.5,9.25,9, 8.75 , 8.5, 8.25, 8...

Step-by-step explanation:

Subtract 0.25 from each to find the next number

8 0

2 years ago

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The position equation for a particle is s of t equals the square root of the quantity t cubed plus 1 where s is measured in feet

vladimir1956 [14]

$\bf s(t)=\sqrt{t^3+1}
\\\\\\
\cfrac{ds}{dt}=\cfrac{1}{2}(t^3+1)^{-\frac{1}{2}}\cdot 3t^2\implies \boxed{\cfrac{ds}{dt}=\cfrac{3t^2}{2\sqrt{t^3+1}}}\leftarrow v(t)
\\\\\\
\cfrac{d^2s}{dt^2}=\cfrac{6t(2\sqrt{t^3+1})-3t^2\left( \frac{3t^2}{\sqrt{t^3+1}} \right)}{(2\sqrt{t^3+1})^2}\implies 
\cfrac{d^2s}{dt^2}=\cfrac{ \frac{6t(2\sqrt{t^3+1})-1}{\sqrt{t^3+1}} }{4(t^3+1)}$

$\bf \cfrac{d^2s}{dt^2}=\cfrac{6t[2(t^3+1)]-1}{4(t^3+1)\sqrt{t^3+1}}\implies 
\boxed{\cfrac{d^2s}{dt^2}=\cfrac{12t^4+12t-1}{4t^3+4\sqrt{t^3+1}}}\leftarrow a(t)\\\\
-------------------------------\\\\a(2)=\cfrac{215~ft^2}{44~sec}$

8 0

2 years ago

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