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1 year ago

Use 3.14 for pi. ROUND YOUR FINAL ANSWER TO THE NEAREST TENTHS PLACE!

Mathematics

2 answers:

Natasha2012 [34]1 year ago

7 0

The area of the given circle is 95 square ft.

Calculation for the Area of the Circle:

It is given in the diagram that,

The diameter of the circle, d = 11 ft.

Now, we know that the radius of a circle is equal to half of the diameter.

Therefore, radius of the given circle, r = 11/2 ft.

The formula for the area of the circle is given as follows,

A = π × r²

Substituting the values, π = 3.1, and r = 11/2, we get,

A = (3.14) × (11/2)²

A = (3.14) × 30.25

A = 94.985

A ≈ 95

Hence, the area of the given circle with diameter 11 ft. comes out to be 95 ft².

Learn more about a circle here:

brainly.com/question/11833983

#SPJ1

ivolga24 [154]1 year ago

4 0

Answer:

379.94

Step-by-step explanation:

3.14 x 11^2 = 379.94

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Evaluate the expression. r = , v = , w = v ⋅ w

FromTheMoon [43]

Answer:

v . w= -13

Step-by-step explanation:

Evaluate the expression: v ⋅ w Given the vectors: r = <8, 1, -6>; v = <6, 7, -3>; w = <-7, 5, 2>

Solution

Given the vectors:

r = <8, 1, -6>

v = <6, 7, -3>

w = <-7, 5, 2>

If you're asking about the dot product.

The dot product is a scalar. It is the sum of the product of the corresponding components.

v.w = (6*-7) + (7*5) + (-3*2)

= -42+35-6

= -13.

6 0

3 years ago

What pattern do you notice in the numbers below? 1 = 1 × 1, and 1 × 1 × 1, and 1 × 1 × 1 64 = 8 × 8, and 4 × 4 × 4 729 = 27 × 27

11111nata11111 [884]

Answer:

<h3>C. They are both perfect squares and perfect cubes.</h3>

Step-by-step explanation:

Perfect squares are numbers that their square root can be found easily without any remainder.

Given the following patterns;

1*1 = 1 and 1*1*1 = 1

It can be seen that 1 is 1 perfect square since 1*1 = 1² = 1

Also 1 is perfect cube since 1*1*1 = 1³ = 1 (cube of the value gives 1)

Similarly for the expression;

8*8 = 64

8² = 64 (since the square of 8 gives 64, then 64 is known to be a perfect square)

Also 4*4*4 = 64

i.e 4³ = 64 (This shows that the cube root of 64 is 4 making it a perfect cube since we can get a whole number for the cube root of 64)

The same is applicable for other expressions 729 = 27 × 27, and 9 × 9 × 9, 4,096 = 64 × 64, and 16 × 16 × 16

This values are easily expressed as a constant multiple of a number showing that they are both perfect squares and perfect cubes.

4 0

3 years ago

HEEEEELLLLPPP ME!!!!!!!!!!!!!!!!!!!!<br> Evaluate u + xy, for u = 2, x = 9, and y = 6.

labwork [276]

Replace u,x and y withyour values
u+xy=2+9*6=2+54=56

5 0

3 years ago

Read 2 more answers

Line segment CD is shown on a coordinate grid:

Jet001 [13]

A) would be your answer hope this helps. :)

4 0

3 years ago

Read 2 more answers

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

3 years ago