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Ivanshal [37]
3 years ago
15

Anyone know this two questions I can't figure this out

Mathematics
2 answers:
ozzi3 years ago
6 0

Answer:

22) x - 46 = 74

23) 20 portions

Step-by-step explanation:

21)

Let x = the number of cars Sara sold.

Jerry sold 46 cars fewer than Sara, so Jerry sold x - 46 cars.

Jerry sold 74 cars, so the equation is

x - 46 = 74

Answer: x - 46 = 74

22)

Let the number of 1/3 lb portions = x

The weight of x number of 1/3-lb portions is 1/3 * x, or simply x/3.

The full bag weighs 6 2/3 lb.

The equation is

x/3 = 6 2/3

Now we solve for x.

x/3 = 20/3

Multiply both sides by 3.

x = 20

Answer: 20 portions

omeli [17]3 years ago
3 0

Answer:

let 's' = Sara's amount

let 's-46' = Jerry's amount

equation:  s - 46 = 74

s = 120

Step-by-step explanation:

2nd question:  \frac{20}{3} ÷ \frac{1}{3} = \frac{20}{3}· \frac{3}{1} = 20

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fgiga [73]

Answer:

100 / 112

Step-by-step explanation:

100 miniuts is a part of 112 adn is not a fraction and it does not say to siplify so you should be good.

6 0
3 years ago
I'm blanking on how to do this, I learned it so long ago, any help would be greatly appreciated. More interested on how to do it
anyanavicka [17]

Answer:

\dfrac{16 y^{22}}{x^{10}z^{10}}

Step-by-step explanation:

Given expression is ,

\sf\longrightarrow \bigg(\dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3}\bigg)^{-2}

This would be simplified using the law of exponents , some of which I will use here are ,

  • (an)^m = a^m n^m
  • \dfrac{a^m}{a^n}=a^{m-n}

  • a^m a^n = a^{m+n}
  • a^{-n} = \dfrac{1}{a^n}

Using the above laws ,

\sf \longrightarrow \bigg[ \dfrac{2}{8} \bigg(\dfrac{x^3}{x^{-2}}\bigg)\bigg(\dfrac{y^{-5}}{y^6}\bigg)\bigg(\dfrac{z^8}{z^3}\bigg)  \bigg]^{-2}

Using the second law mentioned above , we have,

\sf \longrightarrow \bigg[ \dfrac{1}{4}(x^{3+2})(y^{-5-6})(z^{8-3})\bigg]^{-2}

Simplify ,

\sf \longrightarrow \bigg[\dfrac{1}{4} x^5y^{-11}z^5\bigg]^{-2}

Using the first law mentioned above , we have,

\sf \longrightarrow \bigg[ \dfrac{1}{4^{-2}} x^{5(-2)} y^{-11(-2)} z^{5(-2)}\bigg]

Simplify,

\sf \longrightarrow 4^2x^{-10}y^{22} z^{-10}

Finally using the fourth law mentioned above , we have ,

\sf \longrightarrow \boxed{\bf \dfrac{16 y^{22}}{x^{10}z^{10}}}

<h3>Option K is the correct answer.</h3>
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2 years ago
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DerKrebs [107]
A.<span>(x+3)(x+4)=<span>0
x=-3,-4
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</span></span>
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Pie
Well

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