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bulgar [2K]
3 years ago
8

Bc¯¯¯¯¯ is parallel to de¯¯¯¯¯. what is ab? enter your answer in the box.

Mathematics
1 answer:
svetlana [45]3 years ago
7 0
This can help you to understand it

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For F(x) = .01(2) raised to the X find the average rate of change from x=3 to x = 8
erastova [34]
Yes it is going to be 7
7 0
3 years ago
Find the value of 4 + (27 - 12·2) ÷ 2.<br> Answer's are 3.5,5.5 and 19 <br> Which one is it
fredd [130]
You follow the rule PEMDAS parenthesis, exponent, multipy, divide, add, substart. So you would start off by doing 27-12x2, you would do 12x2=24 first then 27-24=3. So then it would be 4+3/2, do 3/2=1.5, then add it 4+1.5=5.5. So the correct answer is 5.5

8 0
3 years ago
Read 2 more answers
A group of boys and girls are trying out for coed hockey. There were 34 children that tried out for the team. The number of girl
Monica [59]

Answer:

Step-by-step explanation:

.

7 0
3 years ago
Raise the quotient of y and 5 to the 2nd power
mojhsa [17]

Answer:

I'm guessing y is numerator and 5 is denominator. so y^2/25

Step-by-step explanation:

so if you have the fraction y/5 and you need to square it or raise to the second power you get (y/5)^2. now using the properties of exponents when you get a fraction and need to raise it to a power you square the denominator and numerator which gets you y^2/25. now let's check if it works we can use 5 for y so (5/5)^2 is equal to 25/25 = 1. or (10/5)^2 which is 2^2 = 4. or 100/25 which is 4. so the equation works for the expression.

5 0
3 years ago
10 POINTS!!! FULL ANSWER IN STEP BY STEP FORMAT!!
stich3 [128]
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).

The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).

b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.

c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.

8 0
3 years ago
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