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rjkz [21]
3 years ago
15

Help ASAP plz and put how to solve it

Mathematics
1 answer:
tatyana61 [14]3 years ago
3 0

Answer: its 1/144

Step-by-step explanation:

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What is the missing term? <br><br> (-14y^2 + 9y^2 -12y +3) + (2y^2 +? -6y -2) = (-3y^2 -15y +1)
harkovskaia [24]

Answer:

3y

Step-by-step explanation:

The missing term is 3y since the following like terms combine:

  • -14y^2+9y^2+2y^2 = -3y^2
  • -12y+-6y = -18y
  • 3+-2 = -1

These terms are the same as the sum listed except -18y.

What must be added to -18y +____=-15y.

-18y+_3y__ =-15y

3y is the solution.


5 0
3 years ago
Average serving size is 500mls which is 1050kj.
astra-53 [7]

Answer:

  2625 kJ

Step-by-step explanation:

We assume the energy is proportional to the volume:

  energy/volume = 1050 kJ/(500 mL) = E/(1250 mL)

Multiply by 1250 mL:

  E = (1250/500)(1050 kJ) = 2625 kJ

The number of kJ in 1.25 L of Lemon Squash is 2625 kJ.

5 0
3 years ago
Find the value of ?<br>bhzkdkekskakak​
azamat
The value of x is 5 since 5^2=25+2=27/3=9
4 0
3 years ago
Please can somebody help me on this one?
TEA [102]

Answer:

  W = 73°

  X = 81°

  Y = 26°

Step-by-step explanation:

We can let W, X, Y represent the measures of the corresponding angles. The problem statement gives us the relations ...

  W = 3Y -5

  X = W +8

Of course, the sum of angles in a triangle is 180°, so we have ...

  W +X +Y = 180

  W +(W +8) +Y = 180 . . . . . substitute for x

  2W +Y = 172 . . . . . . . . . subtract 8, collect terms

  2(3Y -5) +Y = 172 . . . substitute for W

  7Y = 182 . . . . . . . . add 10, collect terms

  Y = 26 . . . . . . . . divide by 7

  W = 3(26) -5 = 73

  X = 73 +8 = 81

The angle measures are (W, X, Y) = (73°, 81°, 26°).

7 0
2 years ago
A motorboat travels due east 6 m/s across a 100m wide river. The current
sladkih [1.3K]

as far as I can make it, the vectors will be like the ones in the picture below, where the green one is the motorboat and the current is in red, and their sum will be the gray one.


since the boat is going 6 m/s in 1 second the motorboat has a magnitude of 6, and since it's due East, is just a horizontal line with an angle of 0°.


the current is due North, therefore has an angle of 90° and since the current is going 3 m/s, so in 1 second it has a magnitude of 3.


\bf \stackrel{\textit{motorboat}}{\begin{cases}x=rcos(\theta )\\\qquad 6cos(0^o)\\\qquad 6\\y=rsin(\theta )\\\qquad 6sin(0^o)\\\qquad 0\end{cases}}\implies \qquad \stackrel{\textit{current}}{\begin{cases}x=3cos(90^o)\\\qquad 0\\y=3sin(90^o)\\\qquad 3\end{cases}}\implies \\\\\\+\implies \implies \\\\\\\measuredangle \theta =tan^{-1}\left(\cfrac{3}{6}  \right)\implies \measuredangle \theta \approx 26.57^o


so, that will be the angle of the resultant vector, now, we know the river is 100 meters wide, so we can use cosine of θ to get the length and therefore the magnitude of the resultant vector. Keeping in mind that the angle of elevation θ is about 26.57° and that the adjacent side is 100.


\bf cos(26.57^o)=\cfrac{\stackrel{adjacent}{100}}{\stackrel{hypotenuse}{h}}\implies h=\cfrac{100}{cos(26.57^o)}\implies h\approx 111.8

7 0
3 years ago
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