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3 years ago

Sam walked 3/5 of a mile to school. He also walked 6/7 of a mile on his way home. How far did Sam walk in all?

2 answers:

7 0

3/5 of 1 mile = 3/5 × 1 = 3 / 5 = 0.6 mile
6/7 of 1 mile = 6/7 × 1 = 6 / 7 = 0.85 mile

total walking distance = 0.6 + 0.85 = 1.45 miles

IF IN FRACTIONS : 3/5 + 6/7
to equalise the denominators multiply 3/5 by 7...........and in 2nd fraction multiply 6/7 by 5

this will become : 21/35 + 30/35
= 51/35 miles

HOPE IT HELPED !!

luda_lava [24]3 years ago

5 0

$\frac{3}{5} + \frac{6}{7}$ you have to get the common denominator which will be:

$\frac{3*7}{5*7} + \frac{6*5}{7*5} = \frac{21}{35} + \frac{30}{35} = \frac{51}{35} or 1 \frac{16}{35} or 1.46$

Hope that helps

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Could anybody please help me on this one? I'm really stuck on this.

san4es73 [151]

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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

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Hello your question is incomplete attached Below is the missing histogram

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