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BlackZzzverrR [31]
3 years ago
7

(5,-2 is on which quardrant

Mathematics
1 answer:
blondinia [14]3 years ago
4 0

Answer:

Quadrant IV, or the fourth quadrant.

Step-by-step explanation:

Plot your points, first plotting 5, then plotting -2.

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Find the surface area of the triangular prism (above) using its net (below).
r-ruslan [8.4K]

Answer:

96 square units

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.  Please have a look at the attached photo.

My answer:

  • The length of the large rectangle is: 5
  • The width of the large rectangle is: 7

=> Area of the large rectangle = length × width  = 7*5 = 35 square units

  • The length of the middle rectangle is: 7
  • The width of the middle rectangle is: 4

=> Area of the middle rectangle = length × width  = 7*4 = 28 square units

  • The length of the small rectangle is: 7
  • The width of the small rectangle is: 3            

=> Area of the small rectangle = length × width  = 7*3 = 21 square units

Area of top and the bottom triangles =2* \frac{1}{2}bh  = 4*3 =12  

Total surface area = 35 + 28 + 21 + 12 = 96 square units

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Shirts are on sale 20% off! The original price of the shirt was $28.50. How
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Answer: im not sure

Step-by-step explanation:

4 0
3 years ago
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What is the quotient оf (x^3 + 3x^2– 4x– 12)-(х^2 + 5х+6)?
DIA [1.3K]

Answer:

The quotient is (x-2)

Step-by-step explanation:

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The solution to this system of equations lies between the x-values -2 and -1.5. At which x-value are the two equations approxima
drek231 [11]
Hello,
\left \{ {{y= \dfrac{1}{x+2} } \atop {y=x^2+2}} \right. \\\\

 \left \{ {{y= \dfrac{1}{x+2} } \atop {\dfrac{1}{x+2}=x^2+2}} \right. \\\\

 \left \{ {{y= \dfrac{1}{x+2} } \atop {x+2= \dfrac{1}{x^2+2} }} \right. \\\\

 \left \{ {{y= \dfrac{1}{x+2} } \atop {x= \dfrac{1}{x^2+2} -2}} \right. \\\\



x=-1.81053571 and y=<span> 5,27803957 </span>

Download xls
3 0
2 years ago
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Calculate the frequency in hertz of photons of light with energy of 8.10 × 10-19 J.
stira [4]
The Planck–Einstein relation E=hv gives that E, the photon energy, is equal to h, Planck's constant, times <span>ν</span>, the frequency.

We can solve the equation in terms of frequency to get
v= \dfrac{E}{h}

We substitute our given energy, 8.10 \times 10^{-19} J, and Planck's constant, 6.626 \times 10^{-34} Js.

v= \dfrac{8.10 \times 10^{-19} J}{6.626 \times 10^{-34} Js}=1.22 \times 10^{15} s^{-1}=1.22 \times 10^{15} Hz
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3 years ago
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