Name at least 4 benchmarks angle measurements.

Tính kết quả \sqrt[3]{1}

vitfil [10]

Answer:

ano po yung question hindi kupo maintindihan pano kupo sa sagutan

6 0

2 years ago

When an amount of heat Q (in kcal) is added to a unit mass (kg) of a

lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71. $\overline {6}$ °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

$\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)$

$\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)$

Therefore, we get;

$\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}$

$\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C } = 71.\overline 6 \, ^{\circ } C/min$

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

$\dfrac{dT}{dt} = 71.\overline 6 \, ^{\circ } C/min$

8 0

2 years ago

WILL GIVE BRAINLEST ! <br> hAVE A GOOD DAy :)!

gayaneshka [121]

Answer:

6. B

7. C

8. B

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Can someone answer part B for me? Thank you and will give brainliest.

frez [133]

Answer:

The answer to part B is B.

Step-by-step explanation:

Could you take another photo so I can see the bottom of the page?

4 0

3 years ago

Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)

zubka84 [21]

$\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } 
$

This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

$\frac{ \sqrt{-49}}{7-2i-4-9i}$

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result:

$\frac{ \sqrt{-49} }{3-11i}$

Now, the $\sqrt{-49}$ must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to $\sqrt{49} \sqrt{-1}$
This means that we get the result 7i for the numerator.

$\frac{7i}{3-11i}$

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: $x^2 - y^2 =(x+y)(x-y)$
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

$\frac{7i (3+11i)}{(3-11i)(3+11i)}$

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how $(3-11i)(3+11i)= 3^2 -(11i)^2$ therefore, we can finally write:

$\frac{7i(3+11i)}{3^2 - (11i)^2 }$

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product $7i(3+11i) = 21i+77i^2$ .
You should know from your classes that i^2 = -1, thefore the numerator simplifies to $-77+21i$
You can do it as a curious thing, but simplifying yields the result:
$\frac{-77+21i}{130}$

7 0

3 years ago