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Alisiya [41]
3 years ago
7

What is the value of fraction 1 over 2x3 3.4y when x = 4 and y = 2? 12.8 14.8 25.8 38.8

Mathematics
1 answer:
just olya [345]3 years ago
7 0
1/2 x^3 + 3.4y when x = 4 and y = 2
1/2(4)^3 + 3.4(2)
1/2 (64) + 6.8
32 + 6.8 = 38.8
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Please helpp due soon!! 10 points :)
zubka84 [21]

Answer:

2.3\:\mathrm{in^3},\\1.3\:\mathrm{fl\:oz}

Step-by-step explanation:

The volume of a cone with radius r and height h is equal to V=\frac{1}{3}\cdot r^2\pi\cdot h.

What we're given:

  • Since radius is half the diameter, r is 1.1
  • Height of 1.8

Substituting given values, we get:

V=\frac{1}{3}\cdot 1.1^2\pi \cdot 1.8=0.726\pi\approx \boxed{2.3\:\mathrm{in^3}}

Convert to fluid ounces using the conversion given in the question:

0.726\pi}\cdot \frac{0.554}{1}\approx \boxed{{1.3\:\mathrm{fl\:oz.}}}

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3 years ago
Samantha and her children went into a movie theater and will buy bags of popcorn and candies. Each bag of popcorn costs $6 and e
Juliette [100K]

Multiply the price of popcorn by the number of bags so 6b.

Multiply the price of candy by the number bought, so 3.25c

Add those together to get total :

6b + 3.25c

Now the most she can spend is 50 so set the equation to less than it equal to what she can spend:

6b + 3.25c <= 50

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3 years ago
Lynn had 90 flyers to hang up around school she gave 12 flyers to each of three volunteers then she took the remaining flyers an
DerKrebs [107]

Answer:

What's the question? If it's how much do each have each of the volunteers have 33 flyers.

7 0
3 years ago
Julian is using a biking app that compares his position to a simulated biker traveling Julian's target speed. When Julian is beh
zepelin [54]

Answer:

\large \boxed{\text{11 km/h}}

Step-by-step explanation:

1. The situation after 15 min:

(a) Distance travelled by simulated biker:

15 min =¼ h  

\text{Distance} = \dfrac{1}{4}\text{ h} \times \dfrac{\text{20 km}}{\text{1 h}} = \text{5 km}

(b) Distance travelled by Julian

Julian is 2¼ km behind  the biker. The distance he has travelled is (5 - 2¼) km

5 - 2¼ = 5 - ⁹/₄ = ²⁰/₄ - ⁹/₄ = ¹¹/₄ = 2¾

Julian has travelled 2¾ km in ¼ h.

2. Julian's average speed

\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{2\frac{3}{4}\text{ km}}{\frac{1}{4}\text{ h}} =\dfrac{11}{4}\text{ km}\times \dfrac{4}{\text{1 h}} = \textbf{11 km/h}\\\\\text{Julian's average speed is $\large \boxed{\textbf{11 km/h}}$}

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