Is 3x-5+y=2y-4 linear if so write the equation in standard form and give the values of A B and C

1 answer:

8 0

Equation in standard form:
Ax+By=C

We have:
3x-5+y=2y-4
3x+y-2y=-4+5
3x-y=1

Therefore:
A=3
B=-1
C=1

Answer: the equation in standard form would be: 3x-y=1; and the values of A, B and C would be: A=3; B=-1 and C=1.

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Find the value of Y.

Deffense [45]

Answer:

5√3

Step-by-step explanation:

The ratio of side lengths in a 30-60-90 triangle is ...

1 : √3 : 2

Multiplying by 5, we have ...

1 : √3 : 2 = x : y : 10 = 5 : 5√3 : 10

The value of y is 5√3.

_____

<em>Additional comment</em>

y is the side adjacent to the 30° angle, so can be found from ...

Cos = Adjacent/Hypotenuse

Adjacent = Cos × Hypotenuse . . . . . multiply by hypotenuse

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y = 5√3

The value of x is 5.

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Suppose a parabola has an axis of symmetry at x=-5, a maximum height of 9, and passes through the point (-7,1). Write the equati

Nutka1998 [239]

the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.

it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.

check the picture below.

so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)

$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9$

$\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill$