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3 years ago

The expression 7d gives the number of days in d weeks . Evaluate 7d for d=12. How many days are in 12 weeks?

Mathematics

2 answers:

e-lub [12.9K]3 years ago

4 0

Answer:

Step-by-step explanation:

RSB [31]3 years ago

3 0

Answer:

84 days are in 12 weeks

Step-by-step explanation:

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2x+4y=16 what is the slope intercept form

Eva8 [605]

Since slope intercept is y=mx+b, you first want to get rid of the 2x by subtracting it from both sides. after, you should end up with 4y=16-2x or 4y=-2x+16, they both are the same, but some teachers might prefer the slope first. after that, you want to get the y alone, so you divide everything by 4. you should end up with the slope intercept form being y=-2/4x+4, simplified to y=-1/2x+4.

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3 years ago

I WILL GIVE BRAINLIEST IF IT IS CORRECT! The Atlantic Ocean covers about 3.2 × 10^7 square miles. The Pacific Ocean covers about

kkurt [141]

Answer:

Step-by-step explanation:

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2 years ago

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The equation of line AB is y = 5x + 1. Write an equation of a line parallel to line AB in slope-intercept form that contains poi

dezoksy [38]

The equation of the line y = 5x + 1. Line that is parallel to the function given should have the same slope as the equation which is 5. We use the point-slope form in order to find the second equation that contains the point (4,5).

y - 5 = 5(x-4)
y = 5x - 20 + 5
y = 5x -15

Therefore, the correct answer is the first option.

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2 years ago

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irina1246 [14]

It is Interest charge

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2 years ago

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

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3 years ago