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labwork [276]
4 years ago
8

What are two numbers multiplied together to get 15 but add to be 12?

Mathematics
1 answer:
KiRa [710]4 years ago
3 0
So x times y=15
x+y=12
solve
x+y=12
subtract x from both sides
y=12-x
subsitute
xy=15
x(12-x)=15
distribute
12x-x^2=15
subtract (12x-x^2) from both sides
0=x^2-12x+15
use quadratic formula which is
x=\frac{ -b+/- \sqrt{b^{2}-4ac} }{2a}
where you have
0=ax^2+bx+c
so
a=1
b=-12
c=15
subsitute
x=\frac{ -(-12)+/- \sqrt{-12^{2}-4(1)(15)} }{2(1)}
x=\frac{ 12+/- \sqrt{144-60} }{2}
x=\frac{ 12+/- \sqrt{84} }{2}
x=\frac{ 12+/- 2\sqrt{21} }{2}
x=6+/- \sqrt{21}
x=6+ \sqrt{21} or 6- \sqrt{21} or aprox
x=10.5826 or 1.41742
the numbers are 10.5826 and 1.41742






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Answer:

6 days

Step-by-step explanation:

The equation for first warehouse would be:

210 + 90x

The equation for 2nd warehouse would be:

180 + 120x

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<em><u>In how many days will the first warehouse have 1.2 times less potatoes than the second warehouse?</u></em>

This means the first warehouse expression TIMES 1.2 would give us 2nd warehouse expression. Equating and solve:

180+120x=1.2(210+90x)\\180+120x=252+108x\\12x=72\\x=\frac{72}{12}\\x=6

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In a two-tailed 2-sample z-test you find a P-Value of 0.0278. At what level of significance would you choose to reject the null
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Answer:

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Step-by-step explanation:

The null hypothesis will be rejected if p-value is less than significance level.

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The fourth graders collected 1.25 pounds more aluminum cans then the fifth graders collected. Select the values that could repre
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Which of the following ratios is equivalent to 5/8 ?
Novosadov [1.4K]

Answer:

Which of the following ratios is equivalent to 5/8 ?

A) 16 : 10

B) 5 to 13  

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<u>D) 15 out of 24</u>

Step-by-step explanation:

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8 0
3 years ago
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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
Oduvanchick [21]

Answer:

  see attached

Step-by-step explanation:

The Pythagorean theorem can be used to find the hypotenuse associated with each pair of legs. That tells you ...

  c² = a² +b² . . . . . legs a, b; hypotenuse c

__

<h3>alternate form of Pythagorean theorem</h3>

For the purpose of this problem, it is convenient to consider a slightly different form of the equation.

For legs √a and √b, the hypotenuse √c is given by ...

  (√c)² = (√a)² +(√b)²

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That is ...

  legs √a, √b ⇒ hypotenuse √(a+b)

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<h3>application to this problem</h3>

Since the legs are (mostly) given in terms of square roots, the value under the radical for the hypotenuse is simply the sum of those:

legs: √1, √2 ⇒ hypotenuse √(1+2) = √3

legs: √2, √3 ⇒ hypotenuse √(2+3) = √5

legs: √5, √3 ⇒ hypotenuse √(5+3) = √8

legs: √5, √1 ⇒ hypotenuse √(5+1) = √6

_____

<em>Additional comment</em>

You may not see the leg lengths given as square roots very often. This is a rather unusual set of problems for hypotenuse length.

6 0
2 years ago
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