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4 years ago

If 2a^2+3^1a-5a^2, then a-a^2=

1 answer:

4 0

$Let's\ 2a^2+3a-5a^2=x\\\\then\\\\2a^2+3a-5a^2=x\\(2a^2-5a^2)+3a=x\\-3a^2+3a=x\\3a-3a^2=x\\3(a-a^2)=x\ \ \ |:3\\\\\boxed{a-a^2=\dfrac{x}{3}}$

$a-a^2\ \text{is 3 times smaller than}\ 2a^2+3a-5a^2$

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In how many distinct ways an the letters of the word Atlanta be arranged

konstantin123 [22]

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3 years ago

What is 1,562 divided by 34

marishachu [46]

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3 years ago

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tatiyna

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Step-by-step explanation:

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3 years ago

Consider h(x)=x^2+8+15. identify its vertex and y-intercept.

OleMash [197]

Answer:

Vertex: (-4, -1)

Y-intercept: (0, 15)

Step-by-step explanation:

Given the quaratic function, h(x) = x² + 8x + 15:

In order to determine the vertex of the given function, we can use the formula, $[x = \frac{-b}{2a}, h(\frac{-b}{2a})]$ .

<h3>Use the equation: $[x = \frac{-b}{2a}, h(\frac{-b}{2a})]$ </h3>

In the quadratic function, h(x) = x² + 8x + 15, where:

a = 1, b = 8, and c = 15:

Substitute the given values for <em>a</em> and <em>b</em> into the equation to solve for the x-coordinate of the vertex.

$x = \frac{-b}{2a}$

$x = \frac{-8}{2(1)}$

x = -4

Subsitute the value of the x-coordinate into the given function to solve for the <u>y-coordinate of the vertex</u>:

h(x) = x² + 8x + 15

h(-4) = (-4)² + 8(-4) + 15

h(-4) = 16 - 32 + 15

h(-4) = -1

Therefore, the vertex of the given function is (-4, -1).

<h3>Solve for the Y-intercept:</h3>

The <u>y-intercept</u> is the point on the graph where it crosse the y-axis. In order to find the y-intercept of the function, set x = 0, and solve for the y-intercept:

h(x) = x² + 8x + 15

h(0) = (0)² + 8(0) + 15

h(0) = 0 + 0 + 15

h(0) = 15

Therefore, the y-intercept of the quadratic function is (0, 15).

5 0

2 years ago

For the equation given below, evaluate y′ at the point (−2,2)<br> xe^y−4y=2x−4−2e^2

jeka94

Implicit differentiation
chain rule is important here

I'll show the steps partially
$e^y+xe^yy'-4y'=2$
$xe^yy'-4y'=2-e^y$
$y'(xe^y-4)=2-e^y$
$y'=\dfrac{2-e^y}{xe^y-4}$
now evaluate for (-2,2)
x=-2 and y=2
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{e^2-2}{2e^2+4}$
that's it, simplest form

3 0

4 years ago