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2 years ago

12

Which point lies on the circle centered at the origin with radius V5?

1 answer:

7nadin3 [17]2 years ago

6 0

Answer:

2, _/21

Step-by-step explanation:

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dalvyx [7]

So,

All we have to do is multiply all three dimensions together.

First, convert all fractions into improper fractions.

$3 \frac{1}{5}--\ \textgreater \ \frac{16}{5}$

$1 \frac{3}{4} --\ \textgreater \ \frac{7}{4}$

$2 \frac{1}{4} --\ \textgreater \ \frac{9}{4}$

Now multiply the fractions together.

<span> $\frac{16}{5}*\frac{7}{4}*\frac{9}{4}= \frac{2*2*2*2*3*3*7}{2*2*2*2*5}$
</span>
$\frac{3*3*7}{5}--\ \textgreater \ \frac{63}{5}\ or\ 12 \frac{3}{5}\ ft.^3$

The correct option is C.

4 0

2 years ago

Alecsey [184]

Answer:

8

Step-by-step explanation:

$6 + 12 - 5 \times 2 + 2(8 \div 4 - 2)$

$6 + 12 - 10 + 2(2 - 2)$

$6 + 12 - 10 + 2 \times 0$

$6 + 12 - 10 + 0$

$= 8$

6 0

2 years ago

14 is 56% of what number

Eddi Din [679]

X= is the number looked for.
56% of x=14
We suggest this equation:
(56/100)x=14
x=(14*100)/56=25

Solution: 25

6 0

2 years ago

Identify: terms like terms, coefficient, constants 10x+3y+8+52

TiliK225 [7]

Like terms are 8 and 52
Coefficients are 10 and 3
Constants are 8 and 52

6 0

3 years ago

How much energy is required to vaporize 185 g of butane at its boiling point? The heat of vaporization for butane is 23.1 kJ/mol

arlik [135]

Answer:

73.5 kJ

Step-by-step explanation:

Mass of butane = 185 g

Heat of vaporization for butane = 23.1 kJ/mol

Molar mass of butane = 58.12 g/mol

Number of moles of butane = $\frac{\text{Mass of butane}}{\text{Molar mass of butane}}=\frac{185}{58.12}=3.18\ moles$

Energy required for burning 185 g of butane = 3.18×23.1 = 73.5 kJ

∴ Energy is required to vaporize 185 g of butane at its boiling point is 73.5 kJ

6 0

2 years ago