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2 years ago

-x + y = -5 Y intersect and x intersect

Mathematics

2 answers:

Gwar [14]2 years ago

6 0

$\\ \sf\longmapsto -x+y=-5$

$\\ \sf\longmapsto x=y+5$

Put x=0

$\\ \sf\longmapsto 0=y+5$

$\\ \sf\longmapsto y=-5$

Y intercept=-5

$\\ \sf\longmapsto x=0+5$

$\\ \sf\longmapsto x=5$

x intercept=5

Furkat [3]2 years ago

5 0

Answer:

(0, - 5) and (5, 0)

Step-by-step explanation:

Given line:

- x + y = - 5

The y-intercept is the point when x = 0:

0 + y = - 5
y = - 5

The point is (0, - 5)

The x-intercept is the point when y = 0:

- x + 0 = - 5
- x = - 5
x = 5

The point is (5, 0)

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2 years ago

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Sunny_sXe [5.5K]

Answer:

Part 1) "a" value is $2$

Part 2) The vertex is the point $(1,8)$

Part 3) The equation of the axis of symmetry is $x=1$

Part 4) The vertex is a minimum

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Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

if a > 0 then the parabola open upward (vertex is a minimum)

if a < 0 then the parabola open downward (vertex is a maximum)

The equation of the axis of symmetry of a vertical parabola is equal to the x-coordinate of the vertex

$x=h$

In this problem we have

$y=2(x-1)^{2}+8$ -----> this is the equation in vertex form of a vertical parabola

The value of $a=2$

a>0 then the parabola open upward (vertex is a minimum)

The vertex is the point $(1,8)$

$(h,k)=(1,8)$

The equation of the axis of symmetry is $x=1$

The equation of a vertical parabola in standard form is equal to

$y=ax^{2}+bx+c$

Convert vertex form in standard form

$y=2(x-1)^{2}+8$

$y=2(x^{2}-2x+1)+8$

$y=2x^{2}-4x+2+8$

$y=2x^{2}-4x+10$

see the attached figure to better understand the problem

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2 years ago

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2 years ago

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Help needed plez Ans in (ii) is 1.5

Deffense [45]

OK first let's check the x=1.5.

$y = \frac x 6 (x^2 - 10)$

$y = 1-2x$

$1-2x = \frac x 6 (x^2 - 10)$

$6-12x = x^3 - 10 x$

$x^3 + 2x-6 = 0$

Oh my, that's called a depressed cubic, no $x^2$ term. There's a formula for these very much like the quadratic formula but you're probably not quite old enough for that. Anyway, $x\approx 1.45616 \approx 1.5$ is a solution, but that's not what they're asking. They are asking us to compare

$x^3 + 2x-6 = 0$

with

$x^3 + bx^2 + cx + d = 0$

and conclude $b=0, c= 2, d=-6$

It turns out we did need all the rest of it. Save those brain cells, there's lots more math coming.

~~~~~~~~~~~~~~

I love it when the student asks for more. Here's the formula for a depressed cubic. I won't derive it here (though I did earlier today, coincidentally, but I'm probably not allowed to link to my Quora answer "what led to the discovery of complex numbers" from here). We use the trick of putting coefficients on the coefficients to avoid fractions.

$x^3 + 3 p x = 2 q$

has solutions

$x = \sqrt[3] { q - \sqrt{p^3 + q^2} } + \sqrt[3] {q + \sqrt{p^3 + q^2} } 

$