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ohaa [14]
3 years ago
10

Evaluate (-1/3)x2/5x(-9)

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
8 0

Answer:

1.2

Let me know if you want me to explain it. :)

lesantik [10]3 years ago
8 0

Answer:

(-1/3x5/2x(-9)

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Compare the functions shown below:
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<span>The function f(x) = x^3 + 2x^2 - 4x - 3 is a polynomial function of degree 3 and hence will have three solutions and hence 3 x-intercepts. The function g(x) has x-intercepts at x = -2, x = -1, x = 1 and x = 3. Hence function g(x) has 4 x-intercepts. The function h(x) has x-intercepts at points (pi over 2, 0) and (3 pi over 2, 0). Hence function h(x) has 2 x-intercepts. Therefore, the function with the most x-intercepts is function g(x) with 4 x-intercepts.</span>
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Jeremy's bank account balance was $35.86 at the beginning of the week. This week, he wrote a $5 check to give to his school's Su
kiruha [24]

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34.61

Step-by-step explanation:

he had 35.86

then gave a 5 dollar check (30.86)

then put in 23.75(54.61)

then took out 20 dollars (34.61)

8 0
3 years ago
What is 42/5 as a mixed number
GarryVolchara [31]
2
8 - Or As a decimal it's 8.4
5
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What is 43/24 as a mixed number?
egoroff_w [7]
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7 0
4 years ago
Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal
babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

Now, the midpoints of these subintervals are:

\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11

Hence:

M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698

(b)

\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557

8 0
3 years ago
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