Answer:
-100
Step-by-step explanation:
1/5 of -125 will be -25.
So, if she deposits 1/5 of what she owes, then she will deposit $25.
Her balance will then become -100 because she paid back 25.
Her new balance is -100
Answer:
x = 54°
y = 66°
Step-by-step explanation:
x + y = 120° (given)
x = y - 12
Then, y + y - 12 = 120°
(y + y) - 12 = 120°
2y - 12 = 120°
2y = 120 + 12 = 132°
y = 132/2 = 66°
x = y - 12 = 66 - 12 = 54°
There you go :)
Answer:
-5
Step-by-step explanation:
-7 - (-2)
-7 + 2
Example: If you owe someone $7, and you give them $2, how much do you still owe? You are owing him $7 so it's a negative and you give him $2 which is a negative as well since you are giving it to him. So, to answer how much do you still owe, you would add $2 to the $7 you owed, so now, you owe him $5, which is a negative since you OWE him.
Hope this helps:)
For this case we must find the product of the following expression:
![\sqrt [3] {5} * \sqrt {2}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B5%7D%20%2A%20%5Csqrt%20%7B2%7D)
By definition of properties of powers and roots we have:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20m%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bm%7D%20%7Bn%7D%7D)
We rewrite the expression using the lowest common index of 6, then:

We rewrite the terms in an equivalent way:

We rewrite the expression using the property mentioned:
![\sqrt [6] {5 ^ 2} * \sqrt [6] {2 ^ 3} =](https://tex.z-dn.net/?f=%5Csqrt%20%5B6%5D%20%7B5%20%5E%202%7D%20%2A%20%5Csqrt%20%5B6%5D%20%7B2%20%5E%203%7D%20%3D)
We combine using the product rule for radicals:
![\sqrt [n] {a} * \sqrt [n] {b} = \sqrt [n] {ab}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%7D%20%2A%20%5Csqrt%20%5Bn%5D%20%7Bb%7D%20%3D%20%5Csqrt%20%5Bn%5D%20%7Bab%7D)
So:
![\sqrt [6] {5 ^ 2 * 2 ^ 3} =\\\sqrt [6] {25 * 8} =\\\sqrt[6]{200}](https://tex.z-dn.net/?f=%5Csqrt%20%5B6%5D%20%7B5%20%5E%202%20%2A%202%20%5E%203%7D%20%3D%5C%5C%5Csqrt%20%5B6%5D%20%7B25%20%2A%208%7D%20%3D%5C%5C%5Csqrt%5B6%5D%7B200%7D)
ANswer:
Option b
The equation y= 2
has one real root and that is x=-1.
What is real roots of the equation?
We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation
-7x+12=0 is solved, the actual roots are 3 and 4.
Here given,
=> y = 2
Take y=0 then,
=> 2
=0
=>
=0
=>(x+1)=0
=> x=-1
Hence the given equation has one real root and that is x=-1.
To learn more about real roots refer the below link
brainly.com/question/24147137
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