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poizon [28]
3 years ago
7

How many solutions does the following equation have? 12z- 6 + 15z = 27z- 5

Mathematics
2 answers:
spin [16.1K]3 years ago
5 0
Zero because you combine 12z and 15z because they are like terms giving you 27z. The equation now looks like 27z-6=27z-5. Subtract 27z on both sides leaving you -6=-5 which is false. So the equation has 0 solutions.
quester [9]3 years ago
3 0

WHO CARE Step-by-step explanation:

i do XD

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7 0
3 years ago
Use substitution to solve the system of equations.<br> x-2y=1<br> 3x-6y=3
Ivahew [28]
Change first equation
x - 2y = 1
add 2y to both side
x = 2y + 1
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3(2y+1) - 6y = 3
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6y + 3 - 6y = 3
y = 0, plug into first equation
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7 0
3 years ago
Multiply by moving decimal point?24.6×100=
Marat540 [252]

Answer:2460

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Please help!! will give brainliest and 20points!!! SUPER EASY PROBLEM
Fiesta28 [93]

Answer:

C. 184 ft²

Step-by-Step Explanation:

The formula for the surface area of a triangular prism is

SA = bh + 2ls + lb; where b = base, h = height, l = length, s = side

First, substitute the known value into the equation:

b = 6 ft

h = 4 ft

l = 10 ft

s = 5 ft

SA = (6)(4) + 2(10)(5) + (10)(6)

Then, all we have to do is simplify:

SA = 24 + 2(50) + 60

SA = 24 + 100 + 60

SA = 184 ft²

Hope this helps!

5 0
3 years ago
Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (x_o, y_o, z_o) and are parallel to \dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

8 0
3 years ago
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