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RoseWind [281]
3 years ago
12

Determine the x-intercept and y-intercept of the following linear equations

Mathematics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

Step-by-step explanation:

the x intercept takes coordinate (x,0)

the y intercept takes coordinate (0,y)

                                         x intercept                        y intercept

40: x+y=5                          (5,0)                                      (0,5)

x=0 , y=5, when y=0 x=5

41: x-y=5                            (5,0)                                      (0,-5)

x=0, -y=5 then y=-5, when y=0, x=5

42: 3x+4y=12                     (4,0)                                      (0,3)

x=0, 4y=12⇒y=12/4⇒y=3

y=0, 3x=12⇒x=12/3⇒x=4

43: 2x+3y=6                      (3,0)                                       (0,2)

x=0 , 3y=6⇒y=6/3=2

y=0 ,3y=6 ⇒ y=6/3 ⇒y=2

44: 3x+4y=-24                   (-8,0)                                      (0,-6)

x=0 , 4y=-24⇒y=-24/4 ⇒ y=-6

y=0 , 3x=-24⇒ x=-24/3⇒x=-8

45: 1/3x+2/3y=5/6            (5/2,0)                                    (0,5/4)

x=0 , 2/3 y=5/6 ⇒12y=15 ⇒ y=15/12 ⇒y=5/4

y=0 , 1/3 x=5/6 ⇒6x=15 ⇒x=15/6 ⇒x=5/2

46: x=-2                             (-2,0)                                   no y intercept

it is vertical line , there is no y intercept

47: y=15                    no x intercept                               (0,15)

horixzontal line, no x intercept

48: 3x-7y=8              (8/3,0)                                             (0,-8/7)

x=0 , -7y=8 ⇒y=-8/7

y=0 , 3x=8 ⇒x=8/3

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\displaystyle\\Answer:\ x=\sqrt{2}\ \ \ \ \ x=-\sqrt{2}\ \ \ \ \ x=\frac{ln5}{2}

Step-by-step explanation:

5x^2-x^2e^{2x}+2e^{2x}=10

Let:\ x^2=t  \ \ \ \ \ e^{2x}=v\\Hence,\\5t-tv+2v^2=10\\5t-tv+2v^2-10=10-10\\5t-tv-10+2v^2=0\\t(5-v)-2(5-v)=0\\(5-v)(t-2)=0\\5-v=0\\5-v+v=0+v\\5=v\\5=e^{2x}

<em>Prologarithmize both parts of the equation:</em>

ln5=lne^{2x}\\ln5=2x*lne\\ln5=2x*1\\ln5=2x\\

<u><em> Divide both parts of the equation by 2:</em></u>

\displaystyle\\x=\frac{ln5}{2}

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In ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45. What is the value of the cosine of ∠X to the nearest hundredth?
kotegsom [21]

The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

Step-by-step explanation:

The given is,

                   In ΔWXY, ∠Y=90°

                        XW = 53

                         YX = 28

                        WY = 45

Step:1

             Ref the attachment,

             Given triangle XWY is right angled triangle.

             Trigonometric ratio's,

                              Cos ∅  = \frac{Adj}{Hyp}    

             For the given attachment, the trigonometric ratio becomes,

                              Cos ∅  = \frac{XY}{XW}.....................................(1)

             Let, ∠X = ∅

             Where, XY = 28

                         XW =  53

             Equation (1) becomes,

                                 Cos ∅  = \frac{28}{53}

                                 Cos ∅ = 0.5283

                                        ∅ = cos^{-1} (0.5283)

                                        ∅ = 58.109°

Result:

          The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

             

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