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rodikova [14]
3 years ago
15

What decimal or fraction goes in the box

Mathematics
1 answer:
garik1379 [7]3 years ago
7 0

Answer:

Your answer would be 7.

Step-by-step explanation:

Hope I helped!!!

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Given: angle 2 and angle 4 are vertical angles Prove: angle 2 congreunt angle 4
7nadin3 [17]

Answer:

Step-by-step explanation:

7 0
2 years ago
In ace g is the centroid and be=12 find bg and ge
Shalnov [3]
For the answer to the question above, uUnder this assumption, BE is a median. Along with any median, the centroid of a triangle is always located 2/3 of the way from the vertex to the midpoint of the opposite side, so the centroid G is 2/3 of the way from E to B. 
So we have GE = (2/3)BE. 

Furthermore, by segment addition, BG + GE = BE. 

Since BE = 9, GE = (2/3)(9) = 6. 
<span>Then BG + 6 = 9, which gives BG = 3</span>
8 0
3 years ago
What is the graph of y = x2 + 6x + 13 ?<br> What does the graph look like
ladessa [460]

Answer:

If I entered it in correctly it would look something like this,,

7 0
3 years ago
Find the Horizontal Tangent for x^3/3+3x^2-16x+9
bazaltina [42]

Answer:

The two horiz. tang. lines here are y = -3 and y = 192.

Step-by-step explanation:

Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function.  Thus, we find f '(x):

f '(x) = x^2 + 6x - 16.  This is the formula for the slope.  We set this = to 0 and determine for which x values the tangent line is horizontal:

f '(x) = x^2 + 6x - 16 = 0.  Use the quadratic formula to determine the roots here:  a = 1; b = 6 and c = -16:  the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:

      -6 plus or minus √100

x = ----------------------------------- = 2 and -8.

                         2

Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3.  So one point of tangency is (2, -3).  Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line:  it is y = -3.

Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8.  The value of x^3/3+3x^2-16x+9 at x = -8 is  192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).


3 0
3 years ago
Solve for the given varrible:<br> A. d=rt Solve for r<br><br> B. I = prt Solve for t
worty [1.4K]
A. d÷t = rt÷t; d/t= r or r= d/t

B.  l÷pr = prt÷pr; l/pr= t or t= l/pr
3 0
3 years ago
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