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I need it ASAP! Melanie had bags of sand for a sandbox. The sandbox needs bags of sand to fill it.

Orlov [11]

<span>Melanie had 2 1/2 bags of sand for a sandbox. The sandbox needs 3 1/4 bags of sand to fill it. Melanie said she still needs 3/4 of a bag to fill it. Is that answer reasonable?
</span><span>
3.25 or 3 and 1/4 - 2.25 or 2 and 1/2= .75 or 3/4</span> so the answer is C.

7 0

3 years ago

Solve the system of equations using any method. 2X – Y + Z = – 2, 6X + 3Y – 4Z = 8, – 3X + 2Y +3Z = – 6

krok68 [10]

Answer:

Z = - 2, Y = 0, X = 0

Step-by-step explanation:

2X – Y + Z = – 2

6X + 3Y – 4Z = 8

– 3X + 2Y +3Z = – 6

isolate "X" in the first equation

2X - Y + Z = - 2 => isolate "2X"

2X = - 2 - Z + Y => divide either side by 2

X = - 1 - Z/2 + Y/2

substitute this value of X in the second two equations (applying the substitution method here)

6(- 1 - Z/2 + Y/2) + 3Y - 4Z = 8,

6Y - 7Z - 6 = 8

- 3(- 1 - Z/2 + Y/2) + 2Y +3Z = - 6,

Y + 9Z + 6/2 = - 6

isolate Y in the second equation (6Y - 7Z - 6 = 8) and substitute in the third equation (Y + 9Z + 6/2 = - 6)

6Y - 7Z - 6 = 8 => isolate 6Y

6Y = 14 + 7Z => divide either side by 6

Y = 7Z + 14/6 => substitute in second equation

7Z + 14/6 + 9Z + 6/2 = - 6 => solve for Z

61Z + 50/12 = - 6, Z = - 2

substitute the value of Z into the equations "Y = 7Z + 14/6" and using the value of Y and Z, substitute into "X = - 1 - Z/2 + Y/2"

Y = (7(- 2) + 14)/6 = 0 / 6

= 0

X = - 1 - (- 2)/2 + 0/2 = - 1 - (-1) + 0

= - 1 + 1 + 0 = 0

6 0

2 years ago

Algebra help please

leva [86]

The value of a would be 4.2, because the negatives cancel out each other.

4 0

3 years ago

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

4 0

3 years ago

Which system of equations is consistent and dependent?

MAXImum [283]

Answer:

Step-by-step explanation:

4x − y = −11

2x + 3y = 5

lets multiply the second equation by -2 and add it to the first:

4x − y = −11

-4x - 6y = -10

------------------

0 - 7y = -21

y = -21/-7

y = 3

now we substitute this result in the first equation to find x:

4x − y = −11

4x - 3 = -11

4x = -8

x = -8/4

x = 2

so the solution is y = 3 and x =2

4x − 9y = −21

−10y = −30

we solve for y

−10y = −30

y = -30/-10

y = 3

and substitute in the first equation:

4x − 9y = −21

4x − 9(3) = −21

4x - 27 = -21

4x = 6

x = 6/4 = 3/2

so the solution is x = 3/2 and y = 3

4x + 3y = 5

2y = −6

we solve for y:

2y = −6

y = -6/2

y = -3

we do substitute in the first equation:

4x + 3y = 5

4x + 3(-3) = 5

4x - 9 = 5

4x = 14

x = 14/4

x = 7/2

so the solution is x = 7/2 and y = -3

7x − 3y = −11

9x = −6

we solve for x:

9x = −6

x = -6/9

x = -2/3

then we substitute in the first equation the result found:

7x − 3y = −11

7(-2/3) − 3y = −11

-14/3 - 3y = -11

we multiply by 3 to eliminate fractions:

-14 - 9y = -33

9y = 19

y = 19/9

so the solution is x = -2/3 and y = 19/9

12x − 3y = −33

14x = −28

we solve for x:

14x = −28

x = -28/14

x = -2

then we substitute in the first equation:

12x − 3y = −33

12(-2) − 3y = −33

-24 - 3y = -33

3y = 9

y = 3

then the solution is x = -2 and y = 3

3 0

3 years ago

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