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2 years ago

Help math math math ASAP ASAP

Mathematics

1 answer:

goldenfox [79]2 years ago

5 0

Answer:

gradient = -1

Step-by-step explanation:

identify the y intercept ( the point where it meet the y- axis ) which denoted by c in the formula y =mx + c

in the graph :the y- intercept is 5 (c)

take the coordinate of any point that passes through the line :- for example (2,3)

in (2,3), the value of x is 2

the value of y is 3

using the formula y = mx + c find m with is the gradient ( unknown )

y = mx + c

3 = m2 + 5

2m + 5 = 3

we sent + 5 to the right and it changes sign to become -5

2m = 3 -5

we substract -5 from 3

2×m = -2

we send 2 to the other side so that m can be alone, it change to become divide since it was multiply

m = -2 ÷ 2

m = -1

the gradient ( unknown ) = -1

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Hello.

The percent decrease in the mass of the pumpkin for this case will be given by:
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After carving it, the pumpkin decreases a percent of 40% in the mass.

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3 years ago

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Give an example of a polynomial that has the asymptotes of x=5 and y=0.5

BaLLatris [955]

(x²+4x+3)/2(x²-10x+25)

the horizontal asymptote when the numerator and the denominator have the same degree (in this case, both of a degree of 2) is ration of the coefficients of the numerator and denominator. In this case, the coefficient for numerator x² is 1, and the coefficient for the denominator 2x² is 2, so the horizontal asymptote is y=1/2=0.5

the vertical asymptote is the x value. the denominator cannot be zero, if x²-10x+25=0, x would be 5, so the vertical asymptote is x=5

this is just one example. There can be others:
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3 years ago

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On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

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How would you determine the axis of symmetry in order to graph x^2 = 8y^2.

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Finding the equation of a parabola is quite difficult but under certain cicumstances we may easily find an equation. Let's place the focus and vertex along the y axis with the vertex at the origin. Suppose the focus is at (0,p). Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The directrix then is the line y=-p. Consider a point P with coordinates (x,y) on the parabola and let Q be the point on the directrix such that the line through PQ is perpendicular to the directrix. The distance PF is equal to the distance PQ. Rather than use the distance formula (which involves square roots) we use the square of the distance formula since it is also true that PF2 = PQ2. We get

(x-0)2+(y-p)2 = (y+p)2+(x-x)2.
x2+(y-p)2 = (y+p)2.If we expand all the terms and simplify, we obtainx2 = 4py.

Although we implied that p was positive in deriving the formula, things work exactly the same if p were negative. That is if the focus lies on the negative y axis and the directrix lies above the x axis the equation of the parabola is

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