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gizmo_the_mogwai [7]
2 years ago
9

Same ratio for 42 and 24 simplify

Mathematics
1 answer:
Ksenya-84 [330]2 years ago
6 0
15:9
hope this helps dont forget to thak me and rate not really it whants 20 charictors but you can be a fan
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Please help me with this.<br> I need a maths genius<br> See pic below.<br> Thanks
adelina 88 [10]

Answer:

a =15

b=70

c=400

d=1500

e=5300

Step-by-step explanation:

it will help u

5 0
3 years ago
What is the variable in the algebraic expression - 9b + 11
vovangra [49]

The variable in this algebraic expression is b

3 0
2 years ago
Read 2 more answers
TRUE OR FALSE!!! PICTURE IS SHOWN
schepotkina [342]
False

distance = square root [(x2−x1)^2+(y2−y1)^2]

5 0
3 years ago
Aj jar of sweets contains 5 yellow sweets, 4 red sweets, 8 green sweets 4 orange sweets and 3 white sweets.
LuckyWell [14K]

Answer:

\frac{9}{24}

Step-by-step explanation:

Add all the sweets together ( that will be your denominator )

5+4+8+4+3= 24

The probability she will pick and yellow and orange ( add those numbers together)

Yellow + orange = Numerator

5+4=9

How you answer should be lay out :

\frac{Orange+ yellow sweets}{Total amount of sweets}

7 0
3 years ago
Read 2 more answers
Suppose that f : [a, b] → [a, b] is continuous. Prove that f has a fixed point. That is, prove that there exists c ∈ [a, b] such
nikklg [1K]

Answer:

Step-by-step explanation:

define the function:

g(x) = f(x) -x

As both f(x) and x are continuous functions, g(x) will also be continuous.

Now, what can we say about g(a) = f(a) -a?

we know that a\leq f(a) \leq b, thus:

a-a\leq f(a)-a \leq b-a\\0 \leq g(a) \leq b-a

thus  g(a) is non-negative.

What about g(b) ? Again we have:

a\leq f(b) \leq b\\a-b \leq f(b) -b \leq  0\\a-b \leq g(b) \leq  0

That means that g(b) is not positive.

Now, we can imagine two cases, either one of g(a) or g(b) is equal to zero, or none of them is. If either of them is equal to zero, we have found a fixed point! In fact, any point c for which g(c)=0 is a fixed point, because:

g(c) = 0 \implies f(c) -c = 0 \implies f(c) = c

Now, if g(a) \neq  0 and g(b) \neq 0, then we have that

g(a) >0 and g(b) < 0. And by Bolzano's theorem we can assert that there must exist a point c between a and b for which g(c)=0. And as we have shown before that point would be a fixed point. This completes the proof.

6 0
3 years ago
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