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3 years ago

Calculate the heat required to melt 1 g of gold at its melting point. The specific heat of fusion for gold is 64.5×103 J/kg.

Mathematics

1 answer:

Galina-37 [17]3 years ago

5 0

Answer:

64.5J

Step-by-step explanation:

Q = ml

Q = (1/1000)kg X 64.5x10^3 J/kg

Q = 64.5 J

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Jenny's bakery sells carrot muffins at 2 dollars each. The electricity to run the over is over 120 dollars per day and the cost

g100num [7]

Answer:

200 muffins

Step-by-step explanation:

The selling price of each muffin = $2

The cost price of each muffin = $1.40

The price of electricity = $120

If the number of muffins that Jenny sells in a day is x, then, Jenny's total cost in a day is:

120 + (1.4 * x) = 120 + 1.4x

and the total sales earnings in a day for x muffins will be:

2 * x = 2x

To break even, the total costs in a day must equal to the total earnings. That is;

$2x = 120 + 1.4x$

Solving this:

$2x - 1.4x = 120\\\\\\0.6x = 120\\\\\\x = 120/0.6\\\\\\x = 200$

She must sell 200 muffins in a day to break even.

7 0

3 years ago

Simplify <img src="https://tex.z-dn.net/?f=%5Csqrt%20-25" id="TexFormula1" title="\sqrt -25" alt="\sqrt -25" align="absmiddle" c

garik1379 [7]

Answer:

5i

Step-by-step explanation:

= $\sqrt{-1 * 25}$
= $\sqrt{-1}\sqrt{25}$
= $\sqrt{-1}$ * 5

Rule: $\sqrt{-1}$ = i

So you get i * 5 = 5i

5 0

3 years ago

Identify an angle adjacent to the given angle.

Step2247 [10]

C.....hlj so yeah because it shows it on the diagram

7 0

3 years ago

Find the measure of angle b.

bija089 [108]

Answer:39

Step-by-step explanation:

90-51=39

4 0

3 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago