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miss Akunina [59]

3 years ago

6

How do i to get 39% of 42

1 answer:

VMariaS [17]3 years ago

5 0

To get this answer you would multiply 42 times 39%

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Compare the perimeters of the two regular figures. Which statement is true?

adell [148]

Answer:

(A). The perimeter of the octagon is greater than that of the hexagon.

Step-by-step explanation:

Since, hexagon consists of 6 sides and 6 angles, thus the measure of one angle of the hexagon will be= $\frac{(n-2){\timeS}180^{\circ}}{6}$

= $\frac{(6-2){\timeS}180^{\circ}}{6}$

= $\frac{(4){\timeS}180^{\circ}}{6}$

= $120^{\circ}$

Now, since MQ is the angle bisector of the one of the angle of the hexagon, therefore ∠QMP=60°.

Now, from ΔQMP. we have

$\frac{MP}{MQ}=cos60^{\circ}$

$MP=\frac{1}{2}$

Thus, the perimeter of the hexagon is:

$P=12{\times}MP$

$P=12{\times}\frac{1}{2}$

$P=6 units$

Thus, the perimeter of hexagon is 6 units.

Also, Since, octagon consists of 8 sides and 8 angles, thus the measure of one angle of the octagon will be= $\frac{(n-2){\timeS}180^{\circ}}{8}$

= $\frac{(8-2){\timeS}180^{\circ}}{8}$

= $\frac{(6){\timeS}180^{\circ}}{8}$

= $135^{\circ}$

Now, since AP is the angle bisector of the one of the angle of the octagon, therefore ${\angle}PAC=cos\frac{135}{2}$ .

From ΔAPC, we have

$AC=cos\frac{135}{2}$

Now, Perimeter of octagon is:

$P=16{\times}cos\frac{135}{2}$

$P=16{\times}0.382$

$P=6.122 units$

Thus, the perimeter of octagon is 6.122 units.

Now, the perimeter of octagon is greater than perimeter of the hexagon, thus option A is correct that is The perimeter of the octagon is greater than that of the hexagon.

4 0

3 years ago

F(x) = 2x2<br> Find f(-6)

Sunny_sXe [5.5K]

Is it 2 times x squared (2x^2?

2 * 6^2
2 * 36
72

3 0

3 years ago

Someone please help!!!!

Ber [7]

Answer:

60

Step-by-step explanation:

3 0

3 years ago

PRECAL TRIG <br> WILL MARK BRAINLIEST IF THEY RESPOND IN 5 MINS

dybincka [34]

Its A

Because

a

is

right

6 0

2 years ago

Primitive function for:<br><br> 1. f(x)=4e2x+e−x, sådan att F(0)=2

Free_Kalibri [48]

The primitive function for $f(x) = 4e^{2x} + e^{-x}$ is given by it's indefinite integral.

To calculate it, recall:

$\frac{d}{dx}e^{ax}=ae^{ax}$

$\int c f(x)dx = c\int f(x)dx$

Let's begin

$\int 4e^{2x}+e^{-x}dx \\4\int e^{2x}dx+ \int e^{-x}dx\\4\frac{e^{2x}}{2}+\frac{e^{-x}}{-1} + c\\$

$F(x) = 2e^{2x}-e^{-x} +c$

To find the costant, we just need to use the fact that $F(0)=2$

$F(0) = 2 \\4e^{0}+e^{0} + c =2\\5+c =2\\c=-3$

Therefore,

$\boxed{F(x) = 2e^{2x} - e^{-x} -3 }$

6 0

2 years ago