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3 years ago

Write the equations included in the same set of related facts as 6 ✖️8=48

Mathematics

2 answers:

makkiz [27]3 years ago

6 0

8×6=48
48÷6=8
48÷8=6

I hope this helped!!!:

Kisachek [45]3 years ago

4 0

8×6=48
48÷8=6
48÷6=8!!!!

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Find the product of (x − 5)^2.

Alex17521 [72]

(x - 5)^2 =
(x - 5)(x - 5) =
x^2 - 5x - 5x + 25 =
x^2 - 10x + 25 <===

8 0

3 years ago

What is the length of the diagonal of a 10 foot by 8 foot rectangle?

zloy xaker [14]

Diagonal of a rectangle formula:

Diagonal = √l² + w²
Diagonal = √10² + 8²
Diagonal = √100 + 64
Diagonal = √164
Diagonal = 12.806 feet.

Or you can use the pythagorean theorem. The diagonal of a rectangle is the hypotenuse of a triangle.

a² + b² = c² ; where a and b represents the lengths of the triangle and c is the hypothenuse. In this case, the lengths are 10 ft and 8ft.

10² + 8² = 100 + 64 = 164 ; This represents c².
c = √c² = √164 = 12.806 ft

6 0

3 years ago

So difficult!!! Someone help

Naya [18.7K]

When two lines cross like this, sum of measures of opposite angles is 180.
this means that:
measure angle AEC + measure angle BED = 180
4x-40 + x+50 = 180
5x +10 = 180
5x = 170
x = 34
therefore:
measure angle AEC = 4x-40 = 4(34)-40 = 96
measure angle BED = x+50 = 34+50 = 84

5 0

3 years ago

Help me plz it’s hard for me lol

slava [35]

Answer:

Step-by-step explanation:

2 *3 sqare root 3 * 12

Solving gives:

6 sqare root 36.

sqare root 36 = 6

2 * 6 * 3

So our answer is 36.

3 0

3 years ago

Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov

mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

$2t=(m+\frac{1}{2} ).\frac{\lambda}{n}$

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

$t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm$

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

$t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm$

Hence the thickness difference should be

$t_1-t_2=131.58-75.19=56.39 \ nm$

7 0

3 years ago