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Lunna [17]
3 years ago
12

Armand ran the 100 yard dash in 17.18 seconds. Arturo's time has an 8 with a value 10 times the value of the 8 in Armand's time.

What could be Arturo time on the 100 yard dash?
Mathematics
1 answer:
frosja888 [35]3 years ago
4 0
8====D

He has a value of 64
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X= 43

That’s the answer
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CAN SOMEONE MULTIPLY THIS PLEASE, ITS URGENT
____ [38]

Let's factorise it :

\: {\qquad  \dashrightarrow \sf   ( {x}^{3}   -  5)(x + 3)  }

\: {\qquad  \dashrightarrow \sf    {x}^{3} (x + 3) + [-5(x + 3)]  }

Using Distributive property we get :

\: {\qquad  \dashrightarrow \sf    {x}^{3} +  {3x}^{3}   + ( - 5x - 15)  }

\: {\qquad  \dashrightarrow \sf    {x}^{3} +  {3x}^{3}   - 5x - 15 }

\: {\qquad  \dashrightarrow \sf    4{x}^{3}  - 5x - 15 }

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Therefore,

\: {\qquad  \dashrightarrow \sf   ( {x}^{3}   -  5)(x + 3)   =4{x}^{3}  - 5x - 15}

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2 years ago
Jasmine is making cookies for everyone in her class and she wants to make sure everyone gets at least one cookie. The recipe cal
andreev551 [17]

Answer:

1.5 or 3/2 teaspoons

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If her class has 36 students, and there are 12 cookies in every batch, then she needs 3 batches to give everyone in her class 1 cookie each. It takes 1/2 teaspoon for every recipe, so she needs 3 of those. 3 × 1/2 = 1.5 (or 3/2)

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Using appreciate property -2/3 × 3/5 + 5/2 - 3/5 × 1/6 ​
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Answer:

The answer for your question is 2.

Step-by-step explanation:

5 0
3 years ago
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Answer both please, the topic is linear inequalities, and please explain it to me.
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<u>Question 1 solution:</u>
You have two unknowns here:

Let the Water current speed = W

Let Rita's average speed = R

We are given <em>two </em>situations, where we can form <em>two equations</em>, and therefore solve for the <em>two unknowns, W, R</em>:

Part 1)  W→ , R←(against current, upstream) 
If Rita is paddling at 2mi/hr against the current, this means that the current is trying to slow her down. If you look at the direction of the water, it is "opposing" Rita, it is "opposite", therefore, our equation must have a negative sign for water<span>:  

</span>R–W=2  - equation 1

Part 2)  W→ , R<span>→</span>(with current)
Therefore, R+W=3  - equation 2

From equation 1, W=R-2, 
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So when W=0 (still), R=5/2mi/hr

Finding the water speed using the same rearranging and substituting process:

1...               R=2+W
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