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3 years ago

12

What is 32.23 + 4x? ASAP

1 answer:

kotykmax [81]3 years ago

8 0

Answer:

cannot be simplified.

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Cos3x+cosx=2cos2xcosx

andrew-mc [135]

$cos(3x)+cos(x)=2\cdot cos\left(\dfrac{3x+x}{2}\right)\cdot cos\left(\dfrac{3x-x}{2}\right)=2\cdot cos(2x)\cdot cos(x).$

Green eyes.

8 0

3 years ago

Need help with these 3

Tamiku [17]

Answer:

1.) x = 110

$\frac{3}{22} =\frac{15}{x}$

So for 3 to become 15 it would have to be multiplied by 5 because 3 x 5 = 15. And if I multiply the numerator by anything I always have to do the same to the denominator. So I multiply 22 x 5 = x. x = 110

$\frac{3 * 5}{22*5} =\frac{15}{110}$

2.) x = 72

$\frac{x}{16} =\frac{9}{2}$

For 16 to become 2 it has to be divided by 8 because 16 ÷ 8 = 2. And I have to do the same to the numerator of course, so i have to find x ÷ 8 = 9. I know the answer by memory, but for some it's easier to rewrite this as 9 × 8 = x. x=72

$\frac{72}{16} =\frac{9}{2}$

3.) c = 24

$\frac{3}{5} =\frac{c}{40}$

For 5 to become 40 it must be multiplied by 8 because 5 × 8 = 40. I have to multiply 8 by the numerator as well and 3 × 8 = 24. c = 24

$\frac{3*8}{5*8} =\frac{24}{40}$

5 0

3 years ago

Read 2 more answers

the intersection of roads m and n creates a 41.4 angle. what are the complementary and supplementary angles of this intersection

MaRussiya [10]

The complementary angle of this intersection is 48.6
The supplementary angle of this intersection is 138.6

7 0

3 years ago

Which polygon appears to be regular?<br><br> - Figure A<br> - Figure B<br> - Figure C<br> - Figure D

Black_prince [1.1K]

Answer:

D

Step-by-step explanation:

Hope this helps! h.a.n.d! (have a nice day.)

5 0

4 years ago

A garden snail travels 0.013 meters per second. If 1 meter = 39.37 inches, how fast is the snail traveling in miles per hour?

Eddi Din [679]

We are given

speed=0.013 m/s

so, we can also write as

$v=0.013\frac{m}{s}$

now, we can change it into miles/hour

so, we know that

1mile =1609.34 m

now, we can find for 1m

$1m=\frac{1}{1609.34} mile$

we know that

$1s=\frac{1}{3600}hr$

we can write our speed as

$v=0.013\frac{1m}{1s}$

we can plug values

$v=0.013\frac{\frac{1}{1609.34} mile}{\frac{1}{3600}hr}$

now, we can simplify it

so, we get

$v=0.02908mi/hr$ ...........Answer

5 0

3 years ago