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4 years ago

I NEED THIS ASAP about the Pythagorean

Mathematics

1 answer:

Natalija [7]4 years ago

8 0

Answer: 24.49 or 24.5 if rounded.

Step-by-step explanation: We have to use the pythagorean theorem of course...

a² + b² = c²

15² + 20² = c²

225 + 400 = c²

600 = c²

$\sqrt{600}$

c = 24.49 (If we round to the nearest tenth then the answer is 24.5)

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Tasya [4]

Answer:

Step-by-step explanation:

we first subtract 5-4 =1 . 62-1+2(8-22)-8
then 8-22 =-14 ( negative 14).
 62-1+2-14-8
then 62-1=61. 61+2=63. 63-14=49. 49-8=41
41 is your final answer

8 0

3 years ago

Write a word phrase that represents x + 5.

baherus [9]

Answer:

5 added to a number, x.

Step-by-step explanation:

I phrased it like other ones I saw before so... :) (Also if you ask someone to write it, I'm sure they'd get x+5)

3 0

4 years ago

Evaluate 11x-4 when x=_4

sveta [45]

I'm not sure if that's x=4 or x= -4, but I'll solve both.
x=4:
11(4) - 4 = 44 - 4 = 40
x = -4:
11(-4) - 4 = -44 -4 = -48
Have an awesome day! :)

4 0

3 years ago

Read 2 more answers

Rapid Rental Car charges $40 rental fee, and $0.25 per mile driven. For the same car,Capital Cars charges $45 for rental fee &am

MArishka [77]

Answer:

100 Miles

Step-by-step explanation:

Let it be for x miles.

40+15+0.25x=45+0.35x

10 =0.1x

x=100

The mile is 45+0.35*100= 80 or 40+15+0.25*100 =80

Have a great Day!

4 0

3 years ago

Lowering powers write in terms of first power of cosine. Cos^6

yaroslaw [1]

The main identity you need is the double angle one for cosine:

$\cos^2x=\dfrac{1+\cos2x}2$

We get

$\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8$

Expand the numerator to apply the identity again:

$\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8$

$\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8$

$\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8$

$\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x$

Finally, make use of the product identity for cosine:

$\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2$

so that ultimately,

$\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x$

$\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x$

4 0

4 years ago