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ddd [48]
3 years ago
15

(349 · 672) + (69 ÷ 49)

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
8 0
234,529.4081632653 is the answer
stich3 [128]3 years ago
6 0

Answer: 234,529.4

Step-by-step explanation:

Multiply 349 * 672 and get 234528

Divide 69/49 and get 1.40816326531

Add: 234528 + 1.40816326531 and get 234529.408163

Round 234529.408163 to the nearest tenth and get 234,529.4

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Does the construction demonstrate how to copy a segment correctly by hand
Lelechka [254]

Answer:

There is not enough context here.

6 0
4 years ago
Alegebra. Identify the vertex and axis of symmetry of each. Please answer and e0lain how and why you got your answers.
Svetlanka [38]

There is no information displayed.

Here is how to do it.

How to calculate axis of symmetry:

Plug your numbers into the axis of symmetry formula. To calculate the axis of symmetry for a 2nd order polynomial in the form ax 2 + bx +c (a parabola), use the basic formula x = -b / 2a. In the example above, a = 2 b = 3, and c = -1. Insert these values into your formula, and you will get:

How to calculate vertex:

The vertex form of a quadratic is given by

y = a(x – h)2 + k, where (h, k) is the vertex.

The "a" in the vertex form is the same "a" as

in y = ax2 + bx + c (that is, both a's have exactly the same value). The sign on "a" tells you whether the quadratic opens up or opens down. Think of it this way: A positive "a" draws a smiley, and a negative "a" draws a frowny. (Yes, it's a silly picture to have in your head, but it makes is very easy to remember how the leading coefficient works.)

In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative.

Suppose that "a" is positive, so a(x – h)2 is zero or positive and, whatever x-value you choose, you're always taking k and adding a(x – h)2 to it. That is, the smallest value y can be is just k; otherwise y will equal k plus something positive. When does y equal only k? When x – h, the squared part, is zero; in other words, when x = h. So the lowest value that y can have, y = k, will only happen if x = h. And the lowest point on a positive quadratic is of course the vertex.

If, on the other hand, you suppose that "a" is negative, the exact same reasoning holds, except that you're always taking k and subtracting the squared part from it, so the highest value y can achieve is y = k at x = h. And the highest point on a negative quadratic is of course the vertex.

8 0
3 years ago
4) What is the difference between -5a + 7 and 3a - 2 ?
bonufazy [111]

Answer:

(3a-2)- (-5a+7.)*

Step-by-step explanation:

The answer is

-2a+5

6 0
3 years ago
Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d
faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

\int\limits^2_0 \ \dfrac{32}{x^2 +4}  \ dx

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint:  \dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1})

k = 4

(b) Now, lets evaluate the same integral using power series.

f(x) = \dfrac{32}{x^2 +4}

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

Step-by-step explanation:

(a)

\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx

= 32 \int\limits^2_0 \dfrac{1}{x+4}\  dx

=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0

= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))

= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))

= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))

= 32 (\dfrac{\pi}{8}-0)

= 32 ( (\dfrac{\pi}{8}))

= 4 \pi

The value of k = 4

(b) \dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...  \ \ \ \ \ (Taylor\ \ Series)

\int\limits^2_0  \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx

S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx +  \dfrac{3}{2^3}\int\limits^2_0 x^4 dx -  \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...

S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...

S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...

S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2)  + \dfrac{3}{9}(2^2) -...

S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...

a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}

6 0
3 years ago
Please help. Sorry to disturb you. Please help me.
Triss [41]
The answer to 3 is C, because 18+15 is 33, and 33 divided into three equal groups would have 11 in each.

The answer to 4 is A, because in 11/3 3 can go into 11 3 times, with 2/3 remaining.
4 0
3 years ago
Read 2 more answers
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