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docker41 [41]
3 years ago
6

Plz help

Mathematics
1 answer:
12345 [234]3 years ago
4 0

Answer:

Why is the picture dark ?

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Does anyone have a word problem idea for this equation 1.25x = 0.75x + 50
PSYCHO15rus [73]

Answer: Keke, do you love me? Are your riding?



Idk either.


Tips: make it something about money, and the k should stand for kittens

5 0
4 years ago
Caroline is making some table decorations.
MrMuchimi

Answer:

Caroline buys 3 packs of candles and 5 packs of holders.

Step-by-step explanation:

1.) First, find multiples of 30 by adding 30 to itself. For example, 30, 60, 90.

2.) Second, find multiples of pakcs of holders until it matches the multiples of candle packs. 18, 36, 54, 72, 90.

3.) then you find that Caroline buys 3 packs of candles and 5 packs of candle holders so she has the same amount of both.

4 0
3 years ago
A coat has been marked down 20% to $26.40 what was it's original value?
Gemiola [76]
20% is equal to 0.2
0.2 times 26.40 is equal to 5.28
26.40 minus 5.28 is equal to 21.12
6 0
3 years ago
find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large
katrin2010 [14]

A generic odd number can be written as

2k+1,\quad k \in \mathbb{Z}

Since there is an odd number every two numbers, three consecutive odd numbers will be

2k+1,\quad 2k+3,\quad 2k+5

Now let's make up the equations: the sum of the first two is

(2k+1)+(2k+3)

And 27 less than 3 times the largest is

3(2k+5)-27

These two must be the same, so we have

(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3

Subtracting 4k and 3 from both sides gives

1=2k \iff k=\dfrac{1}{2}

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0
4 years ago
A flagpole is supported by a wire fastened 60 feet from its base. The wire is longer than the height it reaches on the flagpole.
maxonik [38]

Answer:

135.5ft

Step-by-step explanation:

THIS COMPLETE THE QUESTION

flagpole is supported by a wire fastened 60 feet from its base. The wire is 14 feet longer than the height it reaches on the flagpole. Find the length of the wire.

Let X be the length of the wire

Let y be the height of the pole

X= Y + 14

Y= X -14

We can form a right angle triangle form this, where X denote the Hypotenose, Y is the Opposite, and 60feet is the Adjacent.(CHECK THE ATTACHMENT FOR THE TRIANGLE)

Using Pythagoras theorem

X²= 60² + Y²

X² = 3600 + Y²

if we subsitute for Y we have

X² = 3600 + (X -14)²

X² = 3600 +X² -14X-14X+196

X²-X²+ 28X= 3600+196

28X= 3796

X= 3796/28

=135.5ft

hence length of the wire is 135.5ft

8 0
3 years ago
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