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Nataliya [291]
3 years ago
9

2x - 18x + 14y - 19 Is it equivalent

Mathematics
2 answers:
jek_recluse [69]3 years ago
4 0

Answer:

Step-by-step explanation:

You can simplify this by collecting the like terms. The only like terms are the pair 2x and 18 x

2x-18x = -16x

Now, put them back together,

-16x + 14y-19

Allisa [31]3 years ago
4 0

Answer:

-16x + 14y - 19

Step-by-step explanation:

2x - 18x + 14y - 19

Factor out the x.

x(2-18) + 14y - 19

Solve for the brackets.

x(-16) + 14y - 19

-16x + 14y - 19

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3 years ago
Jody practiced a piano piece for 500 seconds bill practiced a piano piece for 8 minutes who practiced longer explain
Mrac [35]

Answer:

Jody

Step-by-step explanation:

You can either convert 500 seconds to minutes or 8 minutes to seconds to compare. I'll do both.

There is 60 seconds in a minute. To find the total minutes of 500 seconds, divide 500 by 60 → 8.3, which means that Jody practiced for 8.3 minutes.

To find the total seconds of 8 minutes, multiply 60 seconds per minute by 8 minutes → 60 * 8 = 480 seconds which means that Bill practiced for 480 seconds.

Now you can compare. Jody practiced for 500 seconds, or 8.3 minutes. Bill practiced for 480 seconds, or 8 minutes. Jody practiced longer.

7 0
3 years ago
Someone please help!!! This is due tonight!! I’ll give brainliest
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Answer:

($0.5 x 2) + ($1.25 x 3) + ($0.75 x 3) = $7

6 0
3 years ago
Read 2 more answers
A group of 2n people, consisting of n men and n women, are to be independently distributed among m rooms. Each woman chooses roo
vfiekz [6]

Step-by-step explanation:

Assume that

X_i = \left \{ {{1, If , Ith, room, has,exactly, 1,man, and , 1,woman } \atop {0, othewise} \right.

hence,

x = x_1 + x_2+....+x_m

now,

E(x) = E(x_1+x_2+---+x_m)\\\\E(x)=E(x_1)+E(x_2)+---+E(x_m)

attached below is the complete solution

7 0
3 years ago
Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (
natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

I=\int\limits^{}_{} \int\limits^{}_D {x*y}  \, dA =  \int\limits^{}_{} \int\limits^{}_D {x*y}  \, dx*dy

since D is the rectangle such that 0<x<3 , 0<y<3

I=\int\limits^{}_{} \int\limits^{}_D {x*y}  \, dA =  \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y}  \, dx*dy =  \int\limits^{3}_{0} {x}  \, dx\int\limits^{3}_{0} {y}  \, dy  = x^{2} /2*y^{2} /2 =  (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4

4 0
3 years ago
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