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Nezavi [6.7K]
2 years ago
13

What is true about points in Quadrants II and III? A. They all have positive x-coordinates. B. They all have negative x-coordina

tes. C. They all have positive y-coordinates. D. They all have negative y-coordinates.
i dont have much time so pls help
Mathematics
1 answer:
Damm [24]2 years ago
7 0

Answer:

The correct answer is Option B: Quadrants II and III have negative x-coordinates.

Just FYI: Quadrants II and III have both positive and negative y-coordinates.

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Sense it is metal math you do 6 x 200
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2) Solve for x in the following equation: 9x + 17 - 2x -2 = 50 *<br> Your answer
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5

Step-by-step explanation:

7x plus 15 is 50

7x=35

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3 years ago
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A particle begins at point (1, 2) and is moving along the line segment joining (1, 2) to (3, 4). Initially, what is the rate of
Bumek [7]

Answer:

Initially, that is, at (1,2), the rate of increase of the function in the direction given = 0

Hence the function is neither increasing nor decreasing initially in the given direction.

Step-by-step explanation:

f(x,y) = 2x² - y²

Rate of Change of the function = ∇f = (fₓ, fᵧ)

And we're told to find the rate of change in a particular direction = ∇f.û

We first obtain the unit vector in that direction, û

Direction = (3,4) - (1,2) = (2,2)

Uni vector = vector/magnitude

Vector = 2î + 2j

magnitude = √(2² + 2²) = √8 = 2 √2

(2î + 2j)/(2√2) = (1/√2)î + (1/√2)j = û

Unit vector in the direction = (1/√2, 1/√2)

f = 2x² - y²

At (1,2)

fₓ = ∂f/∂x = 4x = 4

fᵧ = ∂f/∂y = -2y = -4

∇f = (4,-4)

∇f.û = (4î - 4j).((1/√2)î + (1/√2)j) = (4/√2) - (4/√2) = 0

If it was positive, then the function is increasing, if it was negative, the function is decreasing. Zero means neither of them.

Hence the function is neither increasing nor decreasing initially in the given direction.

7 0
3 years ago
A car valued at £18000 at the start of 2017, depreciated in value by 5% each year for 3 years. How much did it lose in value ove
Anna11 [10]

<u>Answer:</u>

The amount lost over the 3 years s 2567.25£  

<u>Explanation:</u>

$\mathrm{F}=\mathrm{I} \times\left(1-\left(\frac{r}{100}\right)\right)^{\mathrm{n}}$

where F = final value after n years

I = initial value of the car in 2017 = £18000 (given)

Since the value is depreciated 5% every year for 3 years,

r = percentage rate of depreciation = 5% (given)

n = 3 years

Substituting these values in formula, we get

$\mathrm{F}=18000 \times\left(1-\frac{5}{100}\right)^{3}$

= $18000 \times\left(\frac{95}{100}\right)^{3}$

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Finally 18000-15432.75 = 2567.25£ is the amount lost over this period.  

6 0
3 years ago
Brainliest to the best answer
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Answer:

<u>1</u>

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Is your answer dear .

Hope it helps you:)

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