Question 11a)
We are given side BC equals to side CE and angle CBA equals to angle CED
We also know that angle ACB equals to angle ECD are equal (opposite angles properties)
We have enough information to deduce that triangle ABC and triangle CDE are equal by postulate Angle-Side-Angle (ASA)
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Question 11b)
We are given side AB equal to side ED, side BC equals to side EF, and side AC equals to side DF
We have enough information to deduce that triangle ABC and triangle DEF congruent by postulate Side-Side-Side (SSS)
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Question 11c)
We are given side AC equals to side DF, angle ABC equals to angle DEF, and angle BAC equals to angle EDF
We have enough information to deduce that triangle ABC congruent to triangle DEF by postulate Angle-Side-Angle (ASA)
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Question 11d)
We do not have enough information to tell whether this shape congruent or not
Answer:
Her clay balls weight 0 pound because she doesn't have balls
-13 and -14
They both are consecutive negative integers that multiply to 182.
Answer:
b.
Step-by-step explanation:
We have to look at sign changes in f(x) to determine the possible positive real roots.
There is only one sign change here, between the -8x and the +4. So that means there is only 1 possible real positive root.
Now we have to look at sign changes in f(-x) to determine the possible negative real roots.
There are 3 sign changes here. That means there are either 3 negative roots or 3-2 = 1 negative root. So we have:
1 positive
3 or 1 negative
We need to pair them up now with all the possible combinations.
If we have 1 positive and 1 negative, we have to have 2 imaginary
If we have 1 positive and 3 negative, we have to have 0 imaginary
Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial. This is a 4th degree polynomial, so we will have 4 roots.
6 boxes (add the cumulatively)
each box bas 5 pkgs so 10w,5o
1st box: 10w,5o
2nd box:20w,10o
3rd box:30w,15o
4th box:40w,20o
5th box:50w,25o
6th box:60w,30o
You need a min. of 48w and 27o
