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3 years ago

1.23/16 prosto do matury potrzebuję pomocy!

Mathematics

1 answer:

Anestetic [448]3 years ago

7 0

1.23/16 = 0.076875 . totally answer

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The variable Z is directly proportional to X. When X is 10, Z has the value 190.

irina [24]

Since z is proportion to the x
z = a x
190 = a 10
a = 19

when x = 13
z = a x
z = 19 * 13
z = 247

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3 years ago

Hunter is bisecting the angle shown. When drawing the arcs centered on points P and Q, why must he keep the compass the same wid

S_A_V [24]

Answer:

The arcs are drawn to find a point on the bisecting ray. If the arcs are the same width, it makes sure that they are equidistant from the points on the rays of the angle. This causes the point to be on the bisecting ray.

Step-by-step explanation:

Bisection of an angle implies dividing the angle into two equal parts. The ray that divides the angle is called a bisector.

The hunter should use the same radius or width to draw the two arcs, using points P and Q as the center interchangeably, so that they would intersect at an equidistant point to P and Q. The point of intersection lies on the bisecting ray of the angle.

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3 years ago

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3 years ago

I need Help Fast Asap

Klio2033 [76]

Answer:

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3 years ago

Let f(x)=x^2f ( x ) = x 2. Find the Riemann sum for ff on the interval [0,2][ 0 , 2 ], using 4 subintervals of equal width and t

sladkih [1.3K]

Answer:

$A_L=1.75$

Step-by-step explanation:

We are given:

$f(x)=x^2$

$interval = [a,b] = [0,2]$

Since $n = 4$ ⇒ $\Delta x = \frac{b-a}{n} = \frac{2-0}{4}=\frac{1}{2}$

Riemann sum is area under the function given. And it is asked to find Riemann sum for the left endpoint.

$A_L= \sum\limits^{n}_{i=1}\Delta xf(x_i) = \frac{1}{2}(0^2+(\frac{1}{2})^2+1^2+(\frac{3}{2})^2)=\frac{7}{4}=1.75$

Note:

If it will be asked to find right endpoint too,

$A_R=\sum\limits^{n}_{i=1}\Delta xf(x_i) =\frac{1}{2}((\frac{1}{2})^2+1^2+(\frac{3}{2})^2+2^2)=\frac{15}{4}=3.75$

The average of left and right endpoint Riemann sums will give approximate result of the area under $f(x)=x^2$ and it can be compared with the result of integral of the same function in the interval given.

So, $(A_R+A_L)/2 = (1.75+3.75)/2=2.25$

$\int^2_0x^2dx=x^3/3|^2_0=8/3=2.67$

Result are close but not same, since one is approximate and one is exact; however, by increasing sample rates (subintervals), closer result to the exact value can be found.

3 0

3 years ago