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KonstantinChe [14]
3 years ago
10

Show work please!...

Mathematics
1 answer:
Leni [432]3 years ago
7 0

Answer:what goes around comes around

Step-by-step explanation:

"i need points too thanks"

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Ruth contributes 15% of the total cost of her individual health care. This is a $ 60.00 deduction from each of her biweekly payc
german

Answer:

A.) $10,400

Contribution every payday = $60

Pay periods in a year = 26

Step-by-step explanation:

Assume her salary, that is total cost = y

Percentage deduction 15% of total cost

biweekly paychecks = $60

Therefore, 15% of y = $60

(15/100)y = 60

15y = 6000

y = 400

Total value of contribution:

$400 × number of biweeks in a year

If number of weeks in a year = 52

No of bi-weeks = 52/2 = 26

$400 × 26 = $10400

8 0
3 years ago
A circle is centered at H(4, 0) and has a radius of 10. Where does the point L(-2, 8) lie
Anon25 [30]

Answer:

Step-by-step explanation:

HL=√((-2-4)²+(8-0)²)=√(36+64)=√100=10=radius

so L lies on the circumference of circle.

or eq. of circle is (x-4)²+(y-0)²=10²

(x-4)²+y²=100

Put x=-2,y=8

(-2-4)²+8²=100

36+64=100

or 100=100

which is true.

Hence L lies on the circle.

7 0
4 years ago
Principal, $2000; Annual interest rate, 5.3%; time, 2 years<br> After 2 years?
dangina [55]

Answer:

212 and I= 2000+212

              =2212

Step-by-step explanation:

3 0
3 years ago
PLEASE HELP ME SOLVE THIS WILL GIVE U BRAINIEST!!!!!!!!!!
blsea [12.9K]

Answer: 6.93

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Assume Earth and a cloud layer 770 m above the Earth can be treated as plates of a parallel-plate capacitor.
Travka [436]

Answer:

(a) 1.15×10⁻⁸ F.

(b) 1.5×10⁻⁸ C.

Step-by-step explanation:

(a)

Capacitance of a capacitor: This is the ability for a capacitor to store charge.

The S.I unit of capacitance of a capacitor is Farad (F).

Mathematically, it is represented as

C = ε₀A/d² ................. Equation 1

Where C = Capacitance of the capacitor, ε₀ = permitivity of free space, A = area of the plates, d = distance between the plates.

Given: A = 1×10⁶ m². d = 770 m

Constant: ε₀ = 8.85×10⁻¹² F/m

Substituting these values into equation 1

C = 8.85×10⁻¹²(1×10⁶ )/770

C = 8.85×10⁻⁶/770

C= 0.0115×10⁻⁶

C = 1.15×10⁻⁸ F.

Thus the capacitance of the cloud layer = 1.15×10⁻⁸ F.

(b)

Electric Field Strength: This is the force per unit charge which is exerted at a point in a in a electric field.

It can be represented mathematically as

E = kq/d²........................... Equation 2

Where E = Electric Field Strength, q = charge, d = distance. k = proportionality constant.

Making q the subject of the equation above,

q = Ed²/k ........................... Equation 3.

Given: E = 2×10⁶ N/C, d = 770 m

Constant: k = 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 3

q =  2×10⁶(770)²/(9×10⁹)

q = (2×10⁶)(592900)/(81×10¹⁸)

q = 14639.5×10⁻¹²

q ≈ 1.5×10⁻⁸ C.

Thus the charge the cloud can hold =  1.5×10⁻⁸ C.

3 0
4 years ago
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