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vodomira [7]
2 years ago
14

What is the 5004300 in standard form

Mathematics
2 answers:
Svet_ta [14]2 years ago
6 0

Answer:

In standard form,

5004300. Five Millon four thousand and three hundred.

Nana76 [90]2 years ago
6 0
I don’t know yet hi buggabooo
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can put $11 of his allowance and in his savings account every week call how much money will he have after 15 weeks
dlinn [17]
11 x 15 = 165. he would have 165$
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3 years ago
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PLZ HELP!!!!!!!!!!!!!!!!!!
Inessa05 [86]

Answer:

D.) If soccer balls are sold for $2.05 or $14.61 each, the store will break even but will not make a profit.

Step-by-step explanation:

Let us assume x = selling price of each soccer ball

y = daily profit earned from selling of soccer balls

Given that

Y= -6x^2+100x-180

where,

a = -6

b = 100

c = -180

Now we have to applied the formula which is as follows

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

= \frac{-100\pm\sqrt{100^2-4\times -6\times -180}}{2\times -6}$

= \frac{-100\pm\sqrt{10,000 - 4,320}}{-12}$

= \frac{-100\pm\sqrt{5.680}}{-12}$

= \frac{-100 + 75.3658}{-12}$

= \frac{24.6342}{-12}

x^1 = -2.05285

Now

x^2  = \frac{-100 - 75.3658}{-12}$

= \frac{- 175.3658}{-12}

x^2 = 14.6138

Based on this the option D is most appropriate as per the given situation

4 0
2 years ago
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Complete the inequality. 9% ___ 0.4
Alex73 [517]
9% is less than 0.4. 0.4 is really 4/10 which is 40% 40 is greater than 9 therefore 9% is less than 0.4! So your answer is 9%<0.4! I hope this helped!
3 0
3 years ago
A flat rectangular piece of aluminum has a perimeter of 62 inches. The length is 15
Salsk061 [2.6K]

Answer:

So, the required width of rectangular piece of aluminium is 8 inches

Step-by-step explanation:

We are given:

Perimeter of rectangular piece of aluminium = 62 inches

Let width of rectangular piece of aluminium = w

and length of rectangular piece of aluminium  = w+15

We need to find width i.e value of x

The formula for finding perimeter of rectangle is: Perimeter=2(Length+Width)\\

Now, Putting values in formula for finding Width w:

Perimeter=2(Length+Width)\\\\62=2(w+15+w)\\62=2(2w+15)\\62=4w+30\\62-30=4w\\4w=32\\w=\frac{32}{4}\\w=8

After solving we get the width of rectangular piece :w = 8

So, the required width of rectangular piece of aluminium is 8 inches

6 0
3 years ago
Please help me with this.A rectangular field is
Karolina [17]
First, let's get the perimeter of the rectangle:
P=2W+2L
P=130m+210m
P=340m
Then, let's get the area of the bigger one:
A=WL
A=65m*105m
A=6825m^2

Then let's try using a rectangle with a smaller ratio:
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Then:
A=50m*120m
A=6000m^2

If you used a square:
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A=7225

There you have it. A rectangle with a smaller area with the same perimeter.
What does it show? The smaller the difference you get from width and length, the larger the area is.
5 0
3 years ago
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