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3 years ago

How can a number line be used to explain why 11.6 > 11.3

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

6 0

In a number line its shows u were the spt of the numbers are soo. Lets say the number line goes by 2s. Its starts in 8 soo its 8, 10, 12. Go closer and its 10 , 11 , 12 . closer. 11, 11.1 , 11. 2. Untilit keeps going and u can see that 11.6 is farther to the right so that means that it is greater

Natali [406]3 years ago

3 0

Answer:

We know that,

In the number line, the numbers from the left side to right side are arranged in increasing order,

When we find 11.6 and 11.3 on the number line,

We observed that,

11.6 is to the right of 11.3 on the number line

Therefore, by the above statement,

11.6 is greater than 11.3.

Hence, proved...

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The opposite of -a is always positive. True or false? Answer it FAST please!!

musickatia [10]

Answer:True!

Step-by-step explanation:

The opposite of any negative number (a) will always be positive because -a is the opposite of +a

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3 years ago

Which of the sets of ordered pairs represents a function? A = {(1, -2), (3, -5), (5, 2), (7,5)} B = {(4, 2), (4, -2), (9,3), (9,

Rufina [12.5K]

Answer: only A

all of A’s inputs are different but B’s repeats 4 and 9 making it not a function, so its only A

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3 years ago

Read 2 more answers

Solve the equation using the zero-product property. (2x − 8)(7x + 5) = 0 x = –2 or x = 7 x = 4 or x = x = 4 or x = x = –4 or x =

aliya0001 [1]

$(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}$

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3 years ago

A car requires 22 litres of petrol to travel a distance of 259.6 km.Find

Anon25 [30]

Answer:

Step-by-step explanation:

1. A car requires 22 litres of petrol to travel a distance of 259.6 km

what is the distance that the car can travel on 63 ltr of petrol

22ltr = 259.6km

63ltr=

cross multiply

{63 x 259.6}/22 = 16354.8/22 = 743.4 km

A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 63 ltr of petrol to travel 743.4km

2. To travel a distance of 2013.2 km

we would need to calculate the amount of fuel

A car requires 22 litres of petrol to travel a distance of 259.6 km

what amount of fuel would it require to travel 2013.2km

22ltr = 259.6km

xltr = 2013.2km

x is the value of petrol to cover 2013.2km

cross multiply

(2013.2 x 22)/259.6

44290.4/259.6 = 170.610169492≈170.6 ltr

A car requires 22 litres of petrol to travel a distance of 259.6 km, it would require 170.6 ltr of petrol to travel 2013.2km

if 1ltr is $1.99

170.6 ltr is (170.6 x 1.99)/1 = $339.494≈$339.5

The price of fuel consumed for 2013.2 km at 1 liter of petrol at $1.99 is $339.5

5 0

3 years ago

Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d

Stella [2.4K]

The answer is most likely A.

The integration interval [a, b] is split up into n subintervals of equal length (so each subinterval has width (b - a)/n, same as the coefficient of the sum of y terms) and approximated by the area of n rectangles with base (b - a)/n and height y.

n subintervals require n + 1 points, with

x₀ = a

x₁ = a + (b - a)/n

x₂ = a + 2(b - a)/n

and so on up to the last point x = b. The right endpoints are x₁, x₂, … etc. and the height of each rectangle are the corresponding y 's at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function G(x) continuous* over an interval [a, b] is equal to the slope of the secant line through x = a and x = b, i.e. the value of the difference quotient

(G(b) - G(a) ) / (b - a)

If G(x) happens to be the antiderivative of a function g(x), then this is the same as the average value of g(x) on the same interval,

$g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx$

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [a, b] into n subintervals of equal width (b - a)/n. Over the first subinterval, the area of a trapezoid with "bases" y₀ and y₁ and "height" (b - a)/n is

(y₀ + y₁) (b - a)/n

but y₀ is clearly missing in the sum, and also the next term in the sum would be

(y₁ + y₂) (b - a)/n

the sum of these two areas would reduce to

(b - a)/n = (y₀ + 2 y₁ + y₂)

which would mean all the terms in-between would need to be doubled as well to get

$\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)$

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3 years ago