What symbol do we use for the resultant?

Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t

yanalaym [24]

Let A be the amount of money that Aliya deposited in the savings account. Since A is half as much as money as she invested in a money market account, then the amount that she invested in the market account is 2A.

Express the interest that Aliya earned in terms of A. Set it equal to the amount of $297.60 and then solve for A.

Since the savings account gives 1.9% simple interest, the total amount of interest that she will earn from the savings account is 1.9% of A, which is equal to:

$\frac{1.9}{100}\times A$

Since the money market account gives 3.7% simple interest, the total amount of interest that she will earn from the money market account, is 3.7% of 2A, which is equal to:

$\frac{3.7}{100}\times2A$

Add both interests in terms of A and simplify the expression:

$\begin{gathered} \frac{1.9}{100}\times A+\frac{3.7}{100}\times2A \\ =\frac{1.9}{100}\times A+\frac{7.4}{100}\times A \\ =(\frac{1.9}{100}+\frac{7.4}{100})\times A \\ =\frac{1.9+7.4}{100}\times A \\ =\frac{9.3}{100}\times A \end{gathered}$

The expression (9.3/100)*A represents the total interest after one year. Then:

$\begin{gathered} \frac{9.3}{100}\times A=297.60 \\ \Rightarrow A=\frac{100}{9.3}\times297.60 \\ \Rightarrow A=\frac{100\times297.60}{9.3} \\ \Rightarrow A=\frac{29760}{9.3} \\ \Rightarrow A=3200 \end{gathered}$

Use the value of A to find the amount that was invested in the money market account:

$2A=2\times3200=6400$

Therefore, Aliya deposited 3200 in a savings account and 6400 in a money market account.

3 0

1 year ago

. Which of the following is NOT a property of water? A. Ability to dissolve many substances B. Constant volume upon freezing C.

ss7ja [257]

Answer:

a

Explanation:

water is only able to desolve two substances , salt , sugar,

7 0

3 years ago

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Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

2 years ago

What is the relation between celsius and kelvin

Scilla [17]

Answer:

The Celcius and kelvin scale are related unit for unit. One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point.

7 0

2 years ago

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HELPPP ME PLEASE!!!!

andriy [413]

Answer:

c

Explanation:

3 0

2 years ago

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