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Rufina [12.5K]
2 years ago
5

Find the area of a circle with radius 6 cm .

Mathematics
1 answer:
Ivanshal [37]2 years ago
6 0

Answer:

3.14*6*6=113.6 cm square

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I need help this is dude tomorrow
olganol [36]

Answer:

Negative

Step-by-step explanation:

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2 years ago
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Marta earns a weekly salary of 110$ plus a 6.5% commission on sales at a hobby store. How much would she make in a week if she s
spin [16.1K]

Answer:

503,745

Step-by-step explanation:

4 0
2 years ago
Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-
djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

A(x_{1}, y_{1}})=(-20, 25)

B(x_{2}, y_{2}})=(30, 25)

C(x_{3}, y_{3}})=(30, -10)

D(x_{4}, y_{4}})=(-20, -10)

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}

AB=\sqrt{(30+20)^{2}}

AB=\sqrt{(50)^{2}

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}

BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}

BC=\sqrt{(-35)^{2}}

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}

DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}

DC=\sqrt{(30+20)^{2}}

DC=\sqrt{(50)^{2}}

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}

AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}

AD=\sqrt{(-20+20)^{2}+(-35)^{2}}

AD=\sqrt{(-35)^{2}}

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

7 0
3 years ago
A survey of 37 students at the Wall College of Business showed the following majors: Accounting 9 Finance 6 Economics 3 Manageme
Dmitry [639]

Answer:

10/37.

Step-by-step explanation:

The probability that he or she is a management major is 10/37.

The concept of probability that was used was:

Classical probability: The probability is equal to (number of favourable cases) / (number of total cases). Since the total of students who are a management major is 10 and we have 37 students in total, the probability will be 10/37.

3 0
2 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 [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'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 <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> 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 [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

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)

7 0
3 years ago
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