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Nutka1998 [239]
3 years ago
5

What is 2 - 7 = 5n - 10 solved

Mathematics
1 answer:
Alexeev081 [22]3 years ago
5 0
The answer is 1 :) . i hope i helped ❤️.
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Marty and Ethan both wrote a function, but in different ways.
marishachu [46]

<em><u>Question:</u></em>

Marty and Ethan both wrote a function, but in different ways.

Marty

y+3=1/3(x+9)

Ethan

x y

-4 9.2

-2 9.6

0 10

2 10.4

Whose function has the larger slope?

1. Marty’s with a slope of 2/3

2. Ethan’s with a slope of 2/5

3. Marty’s with a slope of 1/3

4. Ethan’s with a slope of 1/5

<em><u>Answer:</u></em>

Marty’s with a slope of 1/3 has the larger slope

<em><u>Solution:</u></em>

<em><u>Given that Marty equation is:</u></em>

y + 3 = \frac{1}{3}(x+9)

<em><u>The point slope form is given as:</u></em>

y - y_1 = m(x-x_1)

Where, "m" is the slope of line

On comapring both equations,

m = \frac{1}{3}

<em><u>Ethan wrote a function:</u></em>

Consider any two values from the table we have;

(0, 10) and (2, 10.4)

<em><u>The slope is given by formula:</u></em>

m = \frac{y_2-y_1}{x_2-x_1}

From above two points,

(x_1, y_1) = (0, 10)\\\\(x_2, y_2) = (2, 10.4)

Therefore,

m = \frac{10.4-10}{2-0}\\\\m = \frac{0.4}{2} \\\\m = 0.2

Thus we get,

\text{Slope of Ethan} < \text{Slope of Marty}

Therefore, Marty’s with a slope of 1/3  has the larger slope

4 0
3 years ago
Read 2 more answers
Every one reading this please help! I don't understand this and I will give brainliest to the person that helps first with a goo
stira [4]
Since the area of the poster doesn't change by putting it in a frame, we presume the question is asking what the area of the framed poster is.

The length of the poster in its frame is ...
  (frame width on one side) + (poster length) + (frame width on the other side)
  2 in + 32 in + 2 in = 36 in

Likewise, the width of the poster in its frame is ...
  2 in + 24 in + 2 in = 28 in

The area of a rectangle 36 in by 28 in is the product of these dimensions:
  Area = (36 in)×(28 in) = (36×28) in² = 1008 in²
6 0
3 years ago
Read 2 more answers
An object with a starting velocity of 15 m/s accelerates at 3 m/s2. How far does the object travel within 10 seconds?
Anna007 [38]

Answer:  300 m

<u>Step-by-step explanation:</u>

distance(s)=v_it+\dfrac{1}{2}at^2 \quad \text{where}\ v_i \text{is initial velocity, a is acceleration, t is time}

s=(15)(10)+\dfrac{1}{2}(3)(10)^2

  = 150 + 150

  = 300

6 0
3 years ago
Matthew has a 150 page book he has read one third of it how many pages has he read so far
posledela
You take 150/3. The answer is 50.
8 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

<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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