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3 years ago

Compare the values of the underlined digits 6300 and 530

Mathematics

1 answer:

Paha777 [63]3 years ago

4 0

I don't know what's underline but I assume it's the 300 in 6300 and the 30 in 530 so 300>30

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What equation is equivalent to x2 + 20x + 81 = -27? Can you show me how you got the answer please

Alla [95]

The equivalent equation is x² + 20x = -102

<h3>Simplification of quadratic equation</h3>

The equivalent equation can be obtained by factorizing or simplifying the given quadratic equation as shown below;

x² + 20x + 81 = -21

collect similar terms together

x² + 20x + 81 + 21 = 0

x² + 20x + 102 = 0

The equation can also be written as x² + 20x = -102

Thus, the equivalent equation is x² + 20x = -102

Learn more about simplification of quadratic equation here: brainly.com/question/1214333

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2 years ago

Give the slope of each of these items

Whitepunk [10]

Where/what slopes are I talking about?

7 0

3 years ago

9341/2 scientific notation

Darya [45]

Answer: 9.34 x 10^3
Hope this helps

5 0

3 years ago

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

F'(x) = 3x² + 7

and

F (0) = 5

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

The first is done for you, F (0) = 5.

For the remaining b, you can solve for F(x) 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 F (0.1) = 5.701, F (0.2) = 6.408, F (0.5) = 8.625, and F (2.0) = 27.

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

F(x) ≈ L(x) = F (0) + F' (0) (x - 0) = 5 + 7x

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

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

Let y(x) = F(x). Starting with x₀ = 0 and y₀ = F(x₀) = 5, we have

x₁ = x₀ + 0.1 = 0.1

y₁ = y₀ + F'(x₀) (x₁ - x₀) = 5 + 7 (0.1 - 0) → F (0.1) ≈ 5.7

x₂ = x₁ + 0.1 = 0.2

y₂ = y₁ + F'(x₁) (x₂ - x₁) = 5.7 + 7.03 (0.2 - 0.1) → F (0.2) ≈ 6.403

x₃ = x₂ + 0.3 = 0.5

y₃ = y₂ + F'(x₂) (x₃ - x₂) = 6.403 + 7.12 (0.5 - 0.2) → F (0.5) ≈ 8.539

x₄ = x₃ + 1.5 = 2.0

y₄ = y₃ + F'(x₃) (x₄ - x₃) = 8.539 + 7.75 (2.0 - 0.5) → F (2.0) ≈ 20.164

4 0

3 years ago

I need help..................

Angelina_Jolie [31]

1.0 2.39 3.3 4.-17 5.. 15 6. -24 7. 43 8. 39 9. -11 10.-17

5 0

3 years ago

Read 2 more answers