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Jobisdone [24]
3 years ago
14

735+427+527+4777-526+526 times 7 plz it’s due at 1:00

Mathematics
2 answers:
Elza [17]3 years ago
4 0

Answer:

45262

Step-by-step explanation:

Mrrafil [7]3 years ago
3 0

Answer: 45,262

Step-by-step explanation:

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He drove 56 miles because if you divide 168 and 3 it equals 56.

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Hhhhhhhhhheeeeeellllllllllllllllllppppppppppppp
kati45 [8]

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5 0
3 years ago
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How do you do these questions?
Lady bird [3.3K]

Answer:

(e) csc x − cot x − ln(1 + cos x) + C

(c) 0

Step-by-step explanation:

(e) ∫ (1 + sin x) / (1 + cos x) dx

Split the integral.

∫ 1 / (1 + cos x) dx + ∫ sin x / (1 + cos x) dx

Multiply top and bottom of first integral by the conjugate, 1 − cos x.

∫ (1 − cos x) / (1 − cos²x) dx + ∫ sin x / (1 + cos x) dx

Pythagorean identity.

∫ (1 − cos x) / (sin²x) dx + ∫ sin x / (1 + cos x) dx

Divide.

∫ (csc²x − cot x csc x) dx + ∫ sin x / (1 + cos x) dx

Integrate.

csc x − cot x − ln(1 + cos x) + C

(c) ∫₋₇⁷ erf(x) dx

= ∫₋₇⁰ erf(x) dx + ∫₀⁷ erf(x) dx

The error function is odd (erf(-x) = -erf(x)), so:

= -∫₀⁷ erf(x) dx + ∫₀⁷ erf(x) dx

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7 0
3 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
3 years ago
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