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Ray Of Light [21]
2 years ago
7

Work out the equation of the line which has a gradient of 2 and passes through the point (1,4)

Mathematics
1 answer:
Tom [10]2 years ago
8 0

gradient is 2

Step-by-step explanation:

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Find the values of p and q such that 4x^2+12x=4(x+p)^2-q
-Dominant- [34]

Answer:

p = 1.5 and q = 9

Step-by-step explanation:

Expand the right side then compare the coefficients of like terms on both sides, that is

4(x + p)² - q ← expand (x + p)² using FOIL

= 4(x² + 2px + p²) - q ← distribute parenthesis

= 4x² + 8px + 4p² - q

Comparing coefficients of like terms on both sides

8p = 12 ( coefficients of x- terms ) ← divide both sides by 8

p = 1.5

4p² - q = 0 ( constant terms ), that is

4(1.5)² - q = 0

9 - q = 0 ( subtract 9 from both sides )

- q = - 9 ( multiply both sides by - 1 )

q = 9

7 0
3 years ago
What is the fraction 70/75 equivalent to?
ki77a [65]

Answer:14/15

Step-by-step explanation:

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The following stem-and-leaf plot represents the scores earned by Mr. Wood's class on their most recent science test.
rewona [7]

Answer:

i think 10?

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
-3/4(x-3)-5= 1/8 (x-1)
earnstyle [38]
I think it’s 1384939384$3
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2 years ago
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Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
Sedaia [141]

It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt

F(x)=5+(t^3+7t)\bigg|_0^x

F(x)=5+x^3+7x

Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.

On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

4 0
2 years ago
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