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Oduvanchick [21]
3 years ago
12

For each point, identify the axis or quadrant where the point is located.

Mathematics
2 answers:
ser-zykov [4K]3 years ago
8 0
In the point (x, y), if x=0, then the point is on the y axis.
If y=0, the point is on the x axis
If both x and y are positive (greater than 0), the point is in quadrant 1
If x is negative (less than 0) and y is positive (greater than 0), the point is in quadrant 2
If both x and y are negative, the point is in quadrant 3
If x is positive and y is negative, the point is in quadrant 4

Taking (-3, 2), since 3 has a - sign before it (and nothing else), it is negative. Since the x value comes first, this means that x is negative. Since it simply states "2" for the y value, it is positive. As x is negative and y is positive, we know that the point is in quadrant 2.

I challenge you to do this on your own - good luck, and feel free to ask with further questions!
Alona [7]3 years ago
8 0

Answer:

Part 1) (0,16) is located on the Y-axis

Part 2) (2,-12) is located on the IV quadrant

Part 3) (-1,2) is located on the II quadrant

Part 4) (3,0) is located on the X-axis

Part 5) (36.7,3.9) is located on the I quadrant

Part 6) (-\frac{5}{8}, -5\frac{1}{3}) is located on the III quadrant

Step-by-step explanation:

we know that

I Quadrant

The x-coordinate and the y-coordinate are both positive

II Quadrant

The x-coordinate is negative and the y-coordinate is positive

III Quadrant

The x-coordinate and the y-coordinate are both negative

IV Quadrant

The x-coordinate is positive and the y-coordinate is negative

X-axis

The y-coordinate is equal to zero

Y-axis

The x-coordinate is equal to zero

Part 1) point (0,16)

The point is located on the Y-axis

Part 2) point (2,-12)

The point is located on the IV quadrant

Part 3) point (-1,2)

The point is located on the II quadrant

Part 4) point (3,0)

The point is located on the X-axis

Part 5) point (36.7,3.9)

The point is located on the I quadrant

Part 6) point (-\frac{5}{8}, -5\frac{1}{3})

The point is located on the III quadrant

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crimeas [40]

Answer:

The corresponding increase in surface area of the rectangular prism for an increase of 0.9 cm in height is given by

ΔS = [1.8 (L + B)] cm

where L = length of the rectangular prism

B = Breadth of the rectangular prism

Step-by-step explanation:

A rectangular prism is essentially a cuboid.

The surface area of the cuboid is given as

S = 2[LB + LH + BH]

where

L = length of the rectangular prism

B = Breadth of the rectangular prism

H = Height of the rectangular prism

S = 2LB + 2LH + 2BH

Derivative explains that, for small changes in the two quantities,

(∂S/∂H) = (ΔS/ΔH)

ΔS = ΔH × (∂S/∂H)

S = 2LB + 2LH + 2BH

(∂S/∂H) = 2L + 2B = 2(L + B)

ΔH = 0.9 cm

ΔS = ΔH × (∂S/∂H)

ΔS = 0.9 × 2(L + B) = 1.8 (L + B)

A rectangular prism is essentially a cuboid.

The surface area of the cuboid is given as

S = 2[LB + LH + BH]

where

L = length of the rectangular prism

B = Breadth of the rectangular prism

H = Height of the rectangular prism

S = 2LB + 2LH + 2BH

If the height of the rectangular prism increases by 0.9 cm

S(new) = (2×L×B) + 2L(H + 0.9) + 2B(H + 0.9)

S(new) = 2LB + 2LH + 1.8L + 2BH + 1.8B

S(new) = 2LB + 2LH + 2BH + 1.8L + 1.8B

The old surface area, S = 2LB + 2LH + 2BH

Hence, the change or increase in surface area is S(new) - S

[2LB + 2LH + 2BH + 1.8L + 1.8B] - [2LB + 2LH + 2BH] = 1.8L + 1.8B = 1.8 (L + B)

Still the same increase in surface area as obtained by the first method.

Hope this Helps!!!

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Answer:

1.5x+4

Step-by-step explanation:

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Question 1: The y-intercept is where the line crosses the y-axis. The increments of the y-axis is by 20, and the y-intercept is at (0, 20).

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Now find two points:

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