If an atom has 5 protons, 7 neutrons and 5 electrons what is the mass number of the atom?
Please help quick!
No links!

Answers

Answer 1

Answer:

7+5=12

Step-by-step explanation:

To calculate the mass number of an atom, you have to add the protons and the neutrons. We have 7 neutrons and 5 protons, so your mass number would be 12. Let me know if you have any further questions. :)


Related Questions

Today everything at a store is on sale the store offers a 20
% discount the regualr price of a t shirt is 18 what is the discount price

Answers

Answer:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Andrew shovels snow for 4 %2 hours and makes
$27. How much did he make per hour?
And how much does he earn in 8 hours?

Answers

He got paid $54! :)

Write a linear inequality for each graph (back page)

Answers

Answer:

I can't read that...........

(27/8)^1/3×[243/32)^1/5÷(2/3)^2]
Simplify this question sir pleasehelpme​

Answers

Step-by-step explanation:

[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]

[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

We can write as :

27 = 3 × 3 × 3 = 3³

8 = 2 × 2 × 2 = 2³

243 = 3 × 3 × 3 × 3 × 3 = 3⁵

32 = 2 × 2 × 2 ×2 × 2 = 2⁵

[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now, we can write as :

(3³/2³) = (3/2)³

(3⁵/2⁵) = (3/2)⁵

[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now using law of exponent :

[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]

[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]

help me please !!!!

Answers

Answer:

graph X only

Step-by-step explanation:

because with the rate of change it makes a straight line

How many solutions can be found for the system of linear equations represented on the graph?
A) no solution
B) one solution
C) two solutions
D) infinitely many solutions

Answers

Answer:

A) No solution

Step-by-step explanation:

Given the systems of linear equations, y = 2x + 1 and y = 2x - 1:

Both equations in the system have the same slope, m = 2, thus forming parallel lines.  Since their lines are parallel from each other, then it means that their lines will never intersect.

Therefore, the given systems of linear equation is an inconsistent system that has no solution.

Please help. ASAP. Work out, giving your answer in its simplest form:


3 1/2 divided by 2 3/5

Answers

Answer:

26/35

Step-by-step explanation:

1. First to divide the 3 1/2 by 2 3/5 you have to turn them both into improper fractions

First take 3 1/2. You have to multiply the whole number (3) by the denominator (2) and you would get 6. Then you would add then you add the product (6) to the numerator (1) and get 7.

You keep the denominator the same so the improper fraction is 7/2

Do the same thing to 2 3/5 and the improper fraction is 13/5

2. Now we can divide 13/5 by 7/2 using "keep, change, flip"

Keep: 13/5

Change: division to multiplcation

Flip: 7/2 to make 2/7

Your new equation is 13/5 × 2/7. Multiplcation is easy so you just have to multiply staight across: 13 × 2 and 5 × 7 giving you 26/35

If you divide 35 by 26 you will get 1.34 and a bunch of other numbers but I usually stop at two decimal places

hope this helps :)

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