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3 years ago

Kelly plans to take juice on her camping trip. Which will hold more juice, 8 cans or 2 bottles? How much more?

1 answer:

Alex3 years ago

8 0

It depends on how many days you are going to be on your camping trip if you are going to be on the for 2 days you will need the 2 bottles but if you are going on the trip for 8 days you will need the 8 cans. so 8 will work

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What is the solution to this system of equations?

Ede4ka [16]

Answer:

First choice.

Step-by-step explanation:

You could plug in the choices to see which would make all the 3 equations true.

Let's start with (x=2,y=-6,z=1):

2x+y-z=-3

2(2)+-6-1=-3

4-6-1=-3

-2-1=-3

-3=-3 is true so the first choice satisfies the first equation.

5x-2y+2z=24

5(2)-2(-6)+2(1)=24

10+12+2=24

24=24 is true so the first choice satisfies the second equation.

3x-z=5

3(2)-1=5

6-1=5

5=5 is true so the first choice satisfies the third equation.

We don't have to go any further since we found the solution.

---------Another way.

Multiply the first equation by 2 and add equation 1 and equation 2 together.

2(2x+y-z=-3)

4x+2y-2z=-6 is the first equation multiplied by 2.

5x-2y+2z=24

----------------------Add the equations together:

9x+0+0=18

9x=18

Divide both sides by 9:

x=18/9

x=2

Using the third equation along with x=2 we can find z.

3x-z=5 with x=2:

3(2)-z=5

6-z=5

Add z on both sides:

6=5+z

Subtract 5 on both sides:

1=z

Now using the first equation along with 2x+y-z=-3 with x=2 and z=1:

2(2)+y-1=-3

4+y-1=-3

3+y=-3

Subtract 3 on both sides:

y=-6

So the solution is (x=2,y=-6,z=1).

8 0

3 years ago

ASAP WILL MARK BRAINLYEST <br><br><br>answer these two please!

Trava [24]

$19.\\y=3x^2\\a=3,\ b=0,\ c=0\\\\\dfrac{-b}{2a}=\dfrac{-0}{2(3)=0\to x=0\\\\3x^2=0\ \ \ \ |:3\\\\x^2=0\to x=0$

$20.\\y=x^2-x-2\\a=1,\ b=-1,\ c=-2\\\\\dfrac{-b}{2a}=\dfrac{-(-1)}{2(1)}=\dfrac{1}{2}\to x=\dfrac{1}{2}\\\\x^2-x-2=0\\\\x^2-2x+x-2=0\\\\x(x-2)+1(x-2)=0\\\\(x-2)(x+1)=0\iff x-2=0\ \vee\ x+1=0\\\\x=2\ \vee\ x=-1$

3 0

3 years ago

In the diagram, polygon ABCD is flipped over a line of reflection to form a polygon with its vertices at A, B, and D are shown,

Arada [10]

The answer is point C' is (6,2).

8 0

3 years ago

Read 2 more answers

How to find x and y I dont understand

lesantik [10]

68= 4x-2+30 (notice they are vertical angles)
68= 4x+28
40=4x
10=x

68+y= 180 (forms a straight angle)
y=112

Final answer: x=10, y=112

4 0

3 years ago

Read 2 more answers

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