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zaharov [31]
4 years ago
10

662/3% of what number is 320?

Mathematics
1 answer:
sweet [91]4 years ago
4 0
It's similar to example above,I hope you get it.

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ANSWER QUESTION BELLOW!!⇅!!⇅
valkas [14]

Answer:

Option 3. step 1.

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Which equation is quadratic in form?
mixas84 [53]

Answer:

I think it's the first one... but I'm not pretty sure.

4 0
3 years ago
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4 x _ = 10<br> also help me with this plz answer it don't say look it up or use hint
MrMuchimi

Answer:

2.5

Step-by-step explanation:

4 x _ = 10

_ = 10/4

_ = 2.5

If you would like to venture further, you can check out my Instagram page (learntionary) where I post notes and mathematics tips. Thanks!

6 0
3 years ago
Al saves pennines. He agreed to give 6/13 of his pennines to Bev if she would give 6/13 of what she got from Al to Carl and if C
Anuta_ua [19.1K]

Answer:

24,167 pennies

Step-by-step explanation:

Let's see what happens if we say Al had 1000 pennies.

Bev would get 6/13 of 1000 which is  462 pennies. ((6/13)*1000)

Continuing like this Carl would get (6/13)*462 = 213

Dani would get 213(6/13)=98

Now, another way of writing this is 1000*(6/13)*(6/13)*(6/13).  hopefully that makes sense, if not let me know.

Anyway, what is another way of writing this?  If it isn't clear how else can you write 3*3*3*3?  Exponents is the answer.  3*3*3*3=3^4 and (6/13)*(6/13)*(6/13)= (6/13)^3, so you can write the whole equation as 1000(6/13)^3=98

Now, using the information given in the question we know how much is gotten by Dani as well as (6/13)^3, but we don't know how much we start with.  So instead of 1,000 like I showed lets use a variable x.

x*(6/13)^3 = 2376

Now you can use algebra to solve for x

x*(6/13)^3 = 2376 step one let's just figure out what (6/13)^3 is.

x*(216/2197) = 2376  Now here you can either divide by 216/2197 or multiply by 2197 then divide by 216, it's the same thing.

x = 24167

You can double check too with 24167*(6/13)^3

5 0
3 years ago
Use ΔABC to answer the question that follows:
ruslelena [56]

Let's try to render the first part of the proof a bit more legibly.


Point F is a midpoint of Line segment AB

Point E is a midpoint of Line segment AC

Draw Line segment BE

Draw Line segment FC by Construction

Point G is the point of intersection between Line segment BE and Line segment FC Intersecting Lines Postulate

Draw Line segment AG by Construction

Point D is the point of intersection between Line segment AG and Line segment BC Intersecting Lines Postulate

Point H lies on Line segment AG such that Line segment AG ≅ Line segment GH by Construction


OK, now we continue. We need to prove some parallel lines; statement 4 lets us do so.


IV Line segment FG is parallel to line segment BH and Line segment GE is parallel to line segment HC -------- Midsegment Theorem


Now that we've shown some segments parallel we extend that to collinear segments.


III Line segment GC is parallel to line segment BH and Line segment BG is parallel to line segment HC -------- Substitution


We have enough parallel lines to prove a parallelogram


I BGCH is a parallelogram -------- Properties of a Parallelogram (opposite sides are parallel)


Now we draw conclusions from that.


II Line segment BD ≅ Line segment DC -------- Properties of a Parallelogram (diagonals bisect each other)


Answer: IV III I II, second choice


8 0
3 years ago
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