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Viktor [21]
3 years ago
8

In which quadrants does sin and cos have opposite signs.

Mathematics
1 answer:
Romashka [77]3 years ago
8 0
Quadrants 2 and 4 because in quadrant 2, cosine is negative and sin is positive, while in quadrant 4, cosine is positive and sin is negative
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1) Convert fractions into decimals. <br><br>a.34/100<br><br><br>​
Natali5045456 [20]

Answer:

0.34

Step-by-step explanation:

a.

34 / 100 = 0.34

0.34 is the decimal form of the fraction 34 / 100.

8 0
2 years ago
Read 2 more answers
At first it decreased by 60 percent and it increased by 80 percent
storchak [24]

The quantity reported an <em>equivalent net</em> percentage change of 28 percent.

<h3>How to calculate the net change of a quantity in percentages</h3>

In this problem we must determine the <em>simple</em> percentage change equivalent to two <em>consecutive</em> percentual changes. The formula that describes the situation is:

1 + r/100 = (1 - 60/100) · (1 + 80/100)

1 + r/100 = 72/100

r/100 = - 28/100

r = - 28

The quantity reported an <em>equivalent net</em> percentage change of 28 percent.

<h3>Remark</h3>

The statement is incomplete. Complete form is presented below:

A quantity is changing. At first it descreased by 60 percent and it increased by 80 percent. What is net change of the quantity in percentage?

To learn more on percentages: brainly.com/question/13450942

#SPJ1

5 0
1 year ago
Name two pairs of congruent angles in each figure​
Anna007 [38]
EF and GH. those are the right answers
6 0
2 years ago
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The answer to these please!
igomit [66]
\sqrt[4]{(4a^{2})^{6}} = \sqrt[4]{4096a^{12}} = 8a^{3}

\sqrt[7]{(b^{\frac{5}{3}})^{8}} = \sqrt[7]{b^\frac{40}{3}} = b^{\frac{40}{21}} = \sqrt[21]{b^{40}} = \sqrt[21]{b^{21} * b^{19}} = \sqrt[21]{b^{21}} \sqrt[21]{b^{19}} = b\sqrt[21]{b^{19}}

\sqrt[3]{\frac{c^{9}}{c^{3}}} = \sqrt[3]{c^{6}} = c^{2}

\sqrt[3]{27d^{6} * 8d^{-4}} = \sqrt[3]{27d^{6} * \frac{8}{d^{4}}} = \sqrt[3]{216d^{2}} = 6\sqrt[3]{d^{2}} = 6d^{\frac{2}{3}}
4 0
3 years ago
Read 2 more answers
Find derivative of the following problem ?
stepan [7]
Let u = x.lnx, , w= x and t = lnx;  w' =1 ;  t' = 1/x

f(x) = e^(x.lnx) ; f(u) = e^(u); f'(u) = u'.e^(u)

let' find the  derivative u' of u
u = w.t
u'=  w't + t'w;  u' = lnx + x/x = lnx+1

u' = x+1 and f'(u) = ln(x+1).e^(xlnx)

finally the derivative of f(x) =ln(x+1).e^(x.lnx) + 2x
4 0
3 years ago
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