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9966 [12]
4 years ago
15

What point located in quadrant I has an x-value that is 2 units from the origin and a y-value 7 units from the origin?

Mathematics
1 answer:
Lisa [10]4 years ago
4 0

Answer:

The point is P(2,7)

Step-by-step explanation:

We are given the following information in the question:

A point is located in the  quadrant 1 in the following manner:

It has x-value that is 2 units from the origin.

It has a y-value 7 units from the origin.

Thus, the point is (2,7)

The attached image shows the marked point (2,7).

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Two plants, labeled A and B, are both 3 centimeters in height. One month later, plant A is 7 centimeters taller. Plant B is half
Oliga [24]

For this case, we have initially both plants are 3 cm tall.

If a month later the plant A is 7 cm higher, it means that its height will be:

A = 3 + 7 = 10 \ cm

If they tell us that plant B is half the height of plant A, that means:

B = \frac {10} {2} = 5

Answer:

5 centimeters

8 0
3 years ago
Simplify. (3x5+3x2−4x3+x−6)−(x3+4x4+2x2−10x) Enter your answer, in standard form, in the box.
Snezhnost [94]
All so have to do is make sure to convert the negatives and positives of the second group of values since there is a subtraction symbol that you have to distribute to that section.

This would turn the expression into
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Then combine like terms and line all the values in order from greatest exponent value to least to get 
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5 0
4 years ago
Read 2 more answers
Can someone help me with this ????
fenix001 [56]

Answer:

A.

(\frac{3}{3+5})(2 - (-14))+ (-14)

Step-by-step explanation:

Given

Ratio = 3:5

Q = -14

S = 2

Required

Which solution uses (\frac{m}{m+n})(x_2 - x_1)+ x_1

From the given parameters;

m:n = 2:5

m = 2\ and\ n = 5

x_1 = -14\ and\ x_2 = 2

Substitute the above values in (\frac{m}{m+n})(x_2 - x_1)+ x_1

This gives:

(\frac{m}{m+n})(x_2 - x_1)+ x_1

(\frac{3}{3+5})(2 - (-14))+ (-14)

<em>From the list of given options, only option A uses the given form of the formula...</em>

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Q7 Q16.) Convert the polar equation to a rectangular equation.
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In polar coordinates, we have

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