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2 years ago

A restaurant had 82.4 pounds of meat, then they ate 5.5 pounds of it, how much is left

Mathematics

2 answers:

zheka24 [161]2 years ago

5 0

Answer:76.9 pounds

Step-by-step explanation: you subtract 5.5 pounds from the 82.4 pounds to get your answer

Iteru [2.4K]2 years ago

3 0

The answer would be 76.9 because,
-1
-1 12 14
8 2 . 4
- 5 . 5

= 76.9

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A square has an area of 256 square meters. How long is each side? _Meters by _ meters

Lorico [155]

Hello!

First, let's write an equation for the problem.
$\sqrt{256}$

Now, all you have to do is solve it.
$\sqrt{256}$ = 16

Each side is 16 meters.

I hope this helps!

7 0

2 years ago

Justify why a / b x b / c x c /d x d / e is equal to a /e when B C D and E are not zero

Maru [420]

a/b * b/c * c/d * d/e is equal to a/e provided that b, c, d, and e are not zero

PROVE

a/b * b/c * c/d * d/e

= (a/b *b/c) * (c/d * d/e)

= ab/bc * (c/d * d/e)

= a/c * (c/d * d/e)

= a/c * (cd/de)

= a/c * c/e

= ac/ce

= a/e

Therefore, a/b * b/c * c/d * d/e is equal to a/e provided that b, c, d, and e are not zero

7 0

3 years ago

−3x + y = 4, −9x + 5y = −1 find x and y

katrin [286]

Answer:

9x - 3y = -12

-9x +5y = -1

2y = -13

y = -13/2

-3x - 13/2 = 8/2

-3x = 21/2

-6x = 21

x = -21/6 = -7/2

(-7/2, -13/2)

3 0

3 years ago

The video club to which Lin belongs allows her to receive one free DVD for every three DVDS she rents. If she pays $3 for each D

Nesterboy [21]

Answer:

Step-by-step explanation:

Given:

On every 3 DVDs she rents, 1 DVD is received for free

Rent of each DVD = $3

Total amount paid = $114

Now,

The total number of DVDs she took on rent = $\frac{\textup{Total amount paid}}{\textup{Rent of each DVD}}$

The total number of DVDs she took on rent = $\frac{\textup{114}}{\textup{3}}$

The total number of DVDs she took on rent = 38

also,

For every 3 DVDs she gets 1 Free DVD

thus,

For every 1 DVD she gets $\frac{\textup{1}}{\textup{3}}$ free DVD

therefore,

For 38 DVD she gets $\frac{\textup{38}}{\textup{3}}$ free DVD

she gets 12.67 DVDs free

Since,

the DVDs cannot be given in fraction

Hence she will get 12 free DVDs

5 0

3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

7 0

3 years ago