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Triss [41]
3 years ago
5

ILL GIVE BRAINLIYIST

Mathematics
2 answers:
White raven [17]3 years ago
7 0
10°C×9/5+32
= 50°F

so the answer is 50
zimovet [89]3 years ago
6 0
What the first guy said I need those points pls mark me Brainlyist pls illl pray for u
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HELP! ASAP!
masha68 [24]

Step-by-step explanation:

2.

the volume of the box is

length × width × height = 13×13×13 = 13³ = 2197 cm³

the ratio of the density indignation tells us that every cm³ of soil weighs 1.33 g (or every unit of 1.33g soil fits into 1 cm³).

we have 2197 cm³.

their weight is

2197 × 1.33 = 2,922.01 g

so, the filled box clearly exceeds the max. weight of the window ledge, and it is NOT safe to put it there.

b

yes, it would be a little bit less dense.

because every bit of weight you put on top of something increases the pressure on and therefore the density of that something.

6 0
2 years ago
Please show how 1200.00 is an estimate of 31×43?
Vika [28.1K]
Ok so you round 31 to 30 and 43 to 40 and then you multiply 30 by 40 and get 1200. 1200 is the same thing as 1200.00 bc you can take off the zeros after the decimal and it will be the same. Hope this helped and if your still confused you let me know I'll try to explain it more.
4 0
3 years ago
Describe the transformations of f(x) = x3 represented by g(x) = (x - 6) - 2.
devlian [24]

Answer:

Step-by-step explanation:

4 0
3 years ago
How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
lisov135 [29]

Answer:

300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

  • Mass of copper in x pounds of that 15\% copper alloy: (0.15\, x)\; \rm lb.
  • Mass of copper in 700 pounds of that 30\% copper alloy: 700 \times 0.30 = 210\; \rm lb.

Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

4 0
3 years ago
PLEASE HELP ME!!! I ONLY HAVE 2 HOURS!!!
ruslelena [56]
Is this for real? Lol anyways I’m guessing B
6 0
3 years ago
Read 2 more answers
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