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Marina CMI [18]
3 years ago
12

What is the answer to 2a X 3a =

Mathematics
1 answer:
VLD [36.1K]3 years ago
5 0
6a^2
So 6 a squared.
Does this give you what your looking for?

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On a coordinate plane, a line is drawn from point J to point K. Point J is at (negative 3, 1) and point K is at (negative 8, 11)
finlep [7]

Answer:   (-5, 5)

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

J = (-3, 1)    K = (-8, 11)          ratio 2 : 3    --> 2 + 3  = 5 segments

x-distance from J to K: -8 - (-3) = -5 units

y-distance from J to K: 11 - 1 = 10 units

Divide those distance into 5 segments:

x = -5/5  = -1 unit per segment

y = 10/5 = 2 units per segment

The partition is 2 segments from J:

x = -3 +2(-1)  = -5

y = 1 + 2(2)   = 5

The partition is located at (-5, 5)

5 0
4 years ago
Read 2 more answers
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pshichka [43]
Each sentence that has an “as” or “like” is a simile and each that doesn’t is a metaphor. Hope this helps!
5 0
3 years ago
Identify m∠F. PLEASE HELP!!
inysia [295]

Answer:

D. <F = 65 degrees

Step-by-step explanation:

First off, we know that the measures of <F and the angle adjacent to it add to 90 degrees, as indicated by the right angle. They are complementary angles.

The complementary angle of <F has an intercepted arc of 50 degrees. Because the angle is on the opposite end of the circle, it is half of the measure of the arc. Therefore, it is 25 degrees.

Because this angle and <F sum to 90, just subtract 90-25 to find <F.

<F = 65 degrees

7 0
3 years ago
Mr. day for students.<br>​
schepotkina [342]

Answer:

yes

Step-by-step explanation:

5 0
3 years ago
Use the Intermediate Value Theorem to show that the polynomial f(x)=x^3+x^2-2x+5 has a real zero between -3 and -1.
Triss [41]

You need to note first two things.

1. The function f(x) need to be a continuous function on [-3,-1] (indeed f(x) is a polynomial function, is continuous on every real number x)

2. Take the closed interval [-3,-1]

Now, evaluate f(-3) and f(-1).

f(-3) = (-3)^3 + (-3)^2 -2(-3) + 5 = -27 + 9 +6 + 5 = -7 < 0\\f(-1) = (-1)^3 + (-1)^2 -2 (-1) + 5 = -1 + 1 +2 + 5 = 7 >0

So, we have f(-3) < 0  and f(-1). So,  f(-3) < 0 < f(-1), then by the Intermediate Value Theorem, there exist c \in (-3,-1) such that f(c) = 0.

6 0
3 years ago
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