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2 years ago

15

HELP FAST!!!!!!!!! ____ heat is the energy required for 1 kg of a substance to change phase

1 answer:

never [62]2 years ago

5 0

Answer:

The specific latent heat.

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How to do this question plz answer my question step by step plzz plz plz plz

never [62]

Answer:

40°, 95°, 105°, and 160°

Step-by-step explanation:

Let the smallest angle be x.

Measures of the 3 angles can be expressed as:

$x + 55$

$x + 65$

$x + 120$

The sum of all angles in a quadrilateral = 360°.

Therefore, $(x + 55) + (x + 65) + (x + 120) = 360$

Solve for x

$x + 55 + x + 65 + x + 120 = 360$

$3x + 240 = 360$

$3x + 240 - 240 = 360 - 240$

$3x = 120$

$\frac{3x}{3} = \frac{120}{3}$

$x = 40$

The smallest angle = 40°

Plug in the value of x in the earlier stated expressions to find the measure of the other angles:

$x + 55 = 40 + 55 = 95$

$x + 65 = 40 + 65 = 105$

$x + 120 = 40 + 120 = 160$

3 0

3 years ago

Can someone answer this????????????????????????????????/

geniusboy [140]

Answer:

x=4

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I need help with algebra

rosijanka [135]

Answer:

x = -1 and y = 6

Step-by-step explanation:

using x = -b/2a

x = -4 / 2(2)

x = -1

y = 2(-1)² + 4(-1) + 8

y = 2 - 4 + 8

y = 10 - 4

y = 6

4 0

2 years ago

You pick a card at random.

slega [8]

25%

There are 4 different numbers you could draw. Having all 4 cards would be equal to 100%, making each card worth 25%.
25 + 25 + 25 + 25= 100

6 0

3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago