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3 years ago

What is the value of x? 6-2/3(x+5)=4x

Mathematics

1 answer:

gogolik [260]3 years ago

7 0

It was 14

x + 8

= 0

Factorization:

−2

(7x − 4) = 0

Solutions:

x = 8

≈ 0.571429

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horrorfan [7]

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4 years ago

What is the total weight of the bunches of bananas that weigh 1 3\4 pounds

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3 years ago

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marshall27 [118]

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3 years ago

s(t) = 0.29t2 − t + 25 gallons per year (0 ≤ t ≤ 7), where t is time in years since the start of 2007. Use a Riemann sum with n

Sav [38]

Answer: The answer is 300 gallons.

Step-by-step explanation: Riemann sum is a method of calculating the total area under a curve on a graph, which is also known as Integral.

To calculate that area, we divide it into a number of rectangles with one point touching the curve. The curve has a closed interval [a,b] that can be subdivided into n subintervals, each having a width of Δ $x_{k}$ = $\frac{b-a}{n}$

If a function is defined on the closed interval [a,b] and $c_{k}$ is any point in [ $x_{k-1}$ , $x_{k}$ ], then a Riemann Sum is defined as ∑f( $c_{k}$ )Δ $x_{k}$ .

For this question:

Δ $x_{k}$ = $\frac{7-0}{5}$ = 1.4

Now, we have to find s(t) for each valor on the interval:

s(t) = 0.29 $t^{2}$ - t +25

s(0) = 25

s(1) = 24.29

s(2) = 24.16

s(3) = 24.61

s(4) = 25.64

s(5) = 27.25

s(6) = 29.44

s(7) = 32.21

Now, using the formula:

∑f( $c_{k}$ )Δ $x_{k}$ = 1.4(25+24.29+24.16+24.61+25.64+29.44+32.21)

∑f( $c_{k}$ )Δ $x_{k}$ = 1.4(212.6)

∑f( $c_{k}$ )Δ $x_{k}$ ≅ 300

With Riemann Sum, it is estimated the total country's per capita sales of bottled water is 300 gallons.

3 0

3 years ago

If f(x) = x + 4 and g(x)=x^2-1, what is m(g o f)(x)?

Andreas93 [3]

Until now, given a function f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1.

In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and

let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.

Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1).

(i) This type of exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:

8 0

3 years ago

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