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Aliun [14]
3 years ago
13

Fill in the blank: _ _ _ _ _ _ × _ _= 11111111

Mathematics
1 answer:
galina1969 [7]3 years ago
5 0
5555555.5 × 2 = 11111111
Hope it helps!
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Question is on the photo pls help
Bond [772]

Answer:

d)-8

Step-by-step explanation:

  1. -6s-¹t²
  2. -6(3)-¹(-2)²
  3. -6(⅓)(4)
  4. -8
4 0
3 years ago
Each pound of fruit costs $4. Write an expression that shows the total cost of the fruit. Use the variable you identified in que
vovangra [49]

Answer:

$4f

Step-by-step explanation:

Let's assume f represents the number of pounds of fruit.

We need to multiply the number of pounds of fruit by the cost per pound.

This is $f * 4, or in a simpler form, $4f.

6 0
4 years ago
Simplify the expression. Show your work. 22 + (3^2 – 4^2)
irga5000 [103]
<span>22 + (3^2 – 4^2)
= </span><span>22 + (9 – 16)
</span><span>= 22 -7
= 15</span>
6 0
4 years ago
Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz please please please please please please helpppppppppWill give brainliest to right answer
Law Incorporation [45]
A circle with a radius of 4cm sits inside a circle with a radius of 11cm
What is the area of the shaded region?
Round your final answer to the nearest hundredth.

Answer: 329.86 cm
6 0
3 years ago
On Halloween, a man presents a child with a bowl containing eight different pieces of candy. He tells her that she may have thre
likoan [24]

Answer:

56 choices

Step-by-step explanation:

We know that we'll have to solve this problem with a permutation or a combination, but which one do we use? The answer is a combination because the order in which the child picks the candy <u><em>does not</em></u> matter.

To further demonstrate this, imagine I have 4 pieces of candy labeled A, B, C, and D. I could choose A, then C, then B or I could choose C, then B, then A, but in the end, I still have the same pieces, regardless of what order I pick them in. I hope that helps to understand why this problem will be solved with a combination.

Anyways, back to the solving! Remember that the combination formula is

_nC_r=\frac{n!}{r!(n-r)!}, where n is the number of objects in the sample (the number of objects you choose from) and r is the number of objects that are to be chosen.

In this case, n=8 and r=3. Substituting these values into the formula gives us:

_8C_3=\frac{8!}{3!5!}

= \frac{8*7*6*5*4*3*2*1}{3*2*1*5*4*3*2*1} (Expand the factorials)

=\frac{8*7*6}{3*2*1} (Cancel out 5*4*3*2*1)

=\frac{8*7*6}{6} (Evaluate denominator)

=8*7 (Cancel out 6)

=56

Therefore, the child has \bf56 different ways to pick the candies. Hope this helps!

3 0
3 years ago
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