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3 years ago

Determine whether the number is a solution of the given inequality.

Mathematics

1 answer:

myrzilka [38]3 years ago

6 0

The answer to the question is B -2 the other 2 are not correct

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Consider the integral - tan(0) · ln(3 cos(0)) dė:

Lesechka [4]

$\displaystyle \int -\tan(\theta )\cdot \ln(3\cos(\theta )) ~~ d\theta \\\\[-0.35em] ~\dotfill\\\\ u=\ln(3\cos(\theta ))\implies \cfrac{du}{d\theta }=\cfrac{1}{3\cos(\theta )}\cdot -3\sin(\theta ) \\\\\\ \cfrac{du}{d\theta }=-\tan(\theta )\implies \cfrac{du}{-\tan(\theta )}=d\theta \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \int -\tan(\theta )\cdot u\cdot \cfrac{du}{-\tan(\theta )}\implies \int u\cdot du$

4 0

1 year ago

The sum of the numerator and the denominator of a fraction is 216. The fraction is equivalent to $\frac{2}{7}$. What is the valu

il63 [147K]

Answer:

168

Step-by-step explanation:

The fraction is 48/168.

48 + 168 = 216

48/168 = 2/7

The value of the denominator is 168.

We can solve simultaneous equations to get the x and y coordinates.

x + y = 216

x/y = 2/7

x = 48

y = 168

5 0

3 years ago

The cost of 12 pounds of red grapes and 10 pounds of green grapes is

zhenek [66]

Answer:

I believe it is $1.60 a pound but i'm not sure.

Step-by-step explanation:

4 0

3 years ago

a book is 6 inches wide and 9 inches tall A publisher want to scale factor of 1.5 to enlarge a book. what will the area of the f

VladimirAG [237]

Answer:

Step-by-step explanation:

6 inches enlarged by 1.5 = 9 inches

9 inches enlarged by 1.5 = 13.5 inches

Area would be 9x13.5=

121.5 inches for the area

Hope this helps you out.

8 0

4 years ago

FURTHER MATHEMATICS Use determinants to solve the systems of equation:

maw [93]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}2x+1y+2z&=&13\\1x+1y-2z&=&8\\1x+2y+1z&=&11\\\end{array}\right.\\\\\\\Delta=\left| \begin{array}{ccc}2&1&2\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\0&\frac{1}{2}&-3\\0&1&3\end{array}\right| =2*(\frac{3}{2}+3)=9\\\\$

$\Delta_1=\left| \begin{array}{ccc}13&1&2\\8&1&-2\\11&2&1\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\8&1&-2\\\frac{11}{2}& 1&\frac{1}{2}\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\-5&0&-4\\\frac{-5}{2}& 0&\frac{5}{2}\end{array}\right| \\\\=2*(-1)*(\frac{-25}{2}-\frac{20}{2}) =45\\$

$\Delta_2=\left| \begin{array}{ccc}2&13&2\\1&8&-2\\1&11&1\end{array}\right| \\\\\\=\left| \begin{array}{ccc}3&21&0\\3&30&0\\1&11&1\end{array}\right| \\\\\\=1*(90-63) =27\\$

$\Delta_3=\left| \begin{array}{ccc}2&1&13\\1&1&8\\1&2&11\end{array}\right| \\\\\\=\left| \begin{array}{ccc}0&-1&-3\\0&-1&-3\\1&2&11\end{array}\right| \\\\\\=0\\$

$\left\{\begin{array}{ccc}x=\dfrac{\Delta_1}{\Delta}=\dfrac{45}{9}=5\\\\y=\dfrac{\Delta_2}{\Delta}=\dfrac{27}{9}=3\\\\z=\dfrac{\Delta_3}{\Delta}=\dfrac{0}{9}=0\\\end{array}\right.$

8 0

3 years ago