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Klio2033 [76]
2 years ago
6

PLS HELP WILL GIVE BRANLIEST

Mathematics
1 answer:
AnnZ [28]2 years ago
3 0

Answer:

she took 13 classes that were worth 3 credits

Step-by-step explanation:

well we know that some of the classes are worth 3 and some are worth 4 and that all of these together = 59 so your first equation would be

3x+4y=59

x  would be the number of classes worth 3 credits and y would be the number of classes worth 4 credits.

then it says the total number of classes that they took was 18 so we know that

x + y =18.

you want to solve for the number of 3 credit classes so lets go ahead and get rid of the y values. SO you need you multiply the x+y equation by 4

4x+4y=72

then subtract 4x+4y=72  from 3x+4y=59 to get

-x = -13

and then x is a negative so you need to divide -1 from both sides and you get that she took 13 classes that were worth 3 credits.

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The average weight of a newborn panda is 0.2, or 2/10 pound. Between what two integers is the weight of a newborn panda?​
sweet-ann [11.9K]

Answer:

\frac{2}{10} lies between 0 and 1.

Step-by-step explanation:

We must have in mind that given average weight of a newborn panda is a non-integer rational number, such that lies between two integer rational number. In other words, this number must satisfy the following condition:

n < \frac{a}{b} < n +1, n, a, b \in \mathbb{N}_{O}

\mathbb{N}_{O} = \mathbb{N}\,\cup\,{\{0\}}

Which is now developed mathematically to deduce a useful expression to find integers:

1) n < \frac{a}{b} < n +1 Given

2) n \cdot 1 < \frac{a}{b} < n\cdot 1 + 1 Modulative property

3) n \cdot (b\cdot b^{-1}) < a\cdot b^{-1} < n \cdot (b\cdot b^{-1}) + b\cdot b^{-1} Existence of Multiplicative inverse/Definition of division.

4) (n\cdot b)\cdot b^{-1} < a\cdot b^{-1}< [(n+1)\cdot b] \cdot b^{-1} Commutative, Associative and Distributive properties.

5) n\cdot b < a < (n+1)\cdot b Compatibility with Multiplication/Existence of Multiplicative Inverse/Modulative Property/Result.

If we know that a = 2 and b = 10, the following inequation is formed:

10\cdot n < 2 < 10\cdot (n+1)

It is quite evident to conclude that \frac{2}{10} lies between 0 and 1.

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3 years ago
Math question need help with substation please?
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<em>2 solutions</em>

<em>X= 16</em>

<em>X=49</em>

Step-by-step explanation:

<em>original equation</em>

<em>x-11√x+28 = 0</em>

<em>   Isolate</em>

<em>     -11√x = -x-28+0</em>

<em>  Tidy up</em>

<em>     11√x = x+28</em>

<em>  Raise both sides to the second power</em>

<em>     (11√x)2 = (x+28)2</em>

<em>After squaring</em>

<em>     121x = x2+56x+784</em>

<em> Plug in 49 for  x </em>

<em>      11√(49) = (49)+28</em>

<em>Simplify</em>

<em>      11√49 = 77</em>

<em>      Solution checks !!</em>

<em>     Solution is:</em>

     x = 49

<em>Plug in 16 for  x </em>

<em>      11√(16) = (16)+28</em>

<em>Simplify</em>

<em>      11√16 = 44</em>

<em>      Solution checks !!</em>

<em>     Solution is:</em>

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I good accumulator to use would be symbolab.com

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Step-by-step explanation:

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