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rodikova [14]
3 years ago
15

What decimal or fraction goes in the box

Mathematics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

Your answer would be 7.

Step-by-step explanation:

Hope I helped!!!

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Hey can you please help me out on my question posted picture
Nimfa-mama [501]
The equation for a parabola can also be written in the "vertex form":
 y = a (x - h) ^ 2 + k
 In this equation, the vertex of the parabola is the point (h, k).
 We have then:
 y = 2x2 + 16x + 17
 We look for the vertice:
 0 = 4x + 16
 x = -16 / 4
 x = -4
 Substituting:
 y = 2 * (- 4) ^ 2 + 16 * (- 4) +17
 y = -15
 Substituting the given point:
 y = 2 (x +4) ^ 2 - 15
 Answer:
 y = 2 (x +4) ^ 2 - 15
 
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Algebra II Help??? Subtracting Fractions<br><br> (2x)/(y^2-x^2) - (x)/(y-x)
denis23 [38]
Make bottom number same

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\frac{2x}{y^2-x^2}- \frac{(y+x)(x)}{y^2-x^2}=
\frac{2x-x(x+y)}{y^2-x^2}=
\frac{2x-x^2-xy}{y^2-x^2}
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3 years ago
Right Triangle Trig.
d1i1m1o1n [39]

The values are vx = \frac{14\sqrt{3} }{\sqrt{2} }, vw = \frac{14\sqrt{3} }{\sqrt{2} } and m∠x = 45°, for the given right angle diagram.

Step-by-step explanation:

The given is,

                Right angled triangle XVW,

                                     XW = 14\sqrt{3}

                                   m∠V = 90°

                                  m∠W = 45°

Step:1

              Given diagram is right angle triangle,

              Trigonometric ratios for right angle is,

                                 sin ∅ =\frac{Opp}{Hyp}............................(1)

                                  cos ∅ = \frac{Adj}{Hyp} .........................(2)

                                  tan ∅ = \frac{Opp}{Hyp}..........................(3)

Step:2

             For the value of VX,

                                   sin ∅ =\frac{VX}{XW}

            From given,

                               ∅ = 45°

                           XW = 14\sqrt{3}

           Above equation becomes,

                                     sin 45 =\frac{VX}{14\sqrt{3} }

            Where, Sin 45 = \frac{1}{\sqrt{2} },

                                            \frac{1}{\sqrt{2} } = \frac{VX}{14\sqrt{3} }

                                           VX = \frac{14\sqrt{3} }{\sqrt{2} }

Step:3

              For the value of VW,

                                   cos ∅ =\frac{VW}{XW}

            From given,

                               ∅ = 45°

                           XW = 14\sqrt{3}

           Above equation becomes,

                                     cos 45 =\frac{VW}{14\sqrt{3} }

            Where, cos 45 = \frac{1}{\sqrt{2} },

                                            \frac{1}{\sqrt{2} } = \frac{VW}{14\sqrt{3} }

                                           VW = \frac{14\sqrt{3} }{\sqrt{2} }

Step:4

             For the value m∠x = a,

                                      tan a =\frac{VX}{VW}

            From given,

                            VX = \frac{14\sqrt{3} }{\sqrt{2} }

                           VW = \frac{14\sqrt{3} }{\sqrt{2} }

           Above equation becomes,

                                     tan a =\frac{\frac{14\sqrt{3} }{\sqrt{2} } }           {\frac{14\sqrt{3} }{\sqrt{2} } }

                                      tan a = 1

                                             a = tan^{-1} (1)

                                             a = 45°

                                 m∠x = a = 45°

Step:5

            Check for solution,

                         m∠v  = m∠w + m∠x

                                   = 45° + 45°

                            90° =   90°

Result:

            The values are vx = \frac{14\sqrt{3} }{\sqrt{2} }, vw = \frac{14\sqrt{3} }{\sqrt{2} } and m∠x = 45°, for the given right angle diagram.

           

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