w=-6 because you subtract -4 from -10 so the answer is -6
Answer:
yo i do primavara to imma do 3.9
Step-by-step explanation:
oh and also if your grades are bad just do to the dashboard page and highlight you grade percentedge and click inspect if you know and if you see you number pecent on the inspect and right click and click edit as htl and change it to any percent you want then just click out of the inspec option and you will have A S AND B S
Answer:
the hypothesis is that the animal is a poodle. the conclusion is that since poodles are of the canine family, it would be a dog.
The additive inverse of a number is what you can add to that number for the sum to be zero. For real numbers, such as rational numbers, that means the additive inverse can be found by just flipping the sign of the number.
The additive inverse property says if you add a number and its additive inverse, then the sum is zero.
For example, let's use the number 3. The additive inverse of 3 is -3, since 3 + (-3) = 0. This is also true the other way around. The additive inverse of -3 is 3.
When subtracting rational numbers, remember that subtracting is the same as adding a negative! That means 3 - 3 is the same as 3 + (-3) or -3 - (-3) is the same as -3 + 3. Both of these sums involve a number and its additive inverse and they add up to zero.
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For the problem, I don't think you gave me the complete problem, or perhaps there's a typo? Let me know and then I can edit my response to help you out :)
Answer for problem 46 is choice A
Answer for problem 47 is choice B
Answer for problem 48 is choice E
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Work Shown
Problem 46
Equation 1: 3x+y = 17
Equation 2: x+3y = -1
Add equation 1 to equation 2 to get 4x+4y = 16. Divide every term by 4 to get x+y = 4. Then finally multiply both sides by 3 to get 3x+3y = 12
That shows why the answer is choice A
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Problem 47)
If y hours pass by, then y-(2/3)y=y/3 is the time value (2/3)y hours ago
So,
Distance = rate*time
d = r*t
d = x*(y/3)
d = (xy)/3
That's why the answer is choice B
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Problem 48)
Let L1,L2,L3 be the three lists where
L1 = {a1,a2,a3,...,ak} there are k scores here
L2 = {a1,a2,...,a10} there are 10 scores here
L3 = {a11,a12,...,ak} the remaining k-10 scores
S(L1) = sum of the scores in list L1
M(L1) = mean of L1 = 20 = S(L1)/k
M(L2) = mean of L2 = 15 = S(L2)/10
S(L1) = 20k
S(L2) = 150
S(L1) = S(L2)+S(L3)
M(L1) = [S(L2)+S(L3)]/k
20 = [150+S(L3)]/k
20k = 150+S(L3)
S(L3) = 20k-150
M(L3) = [S(L3)]/(k-10)
M(L3) = (20k-150)/(k-10)
So that shows why the answer is choice E
