23y=6 what is y equal to

B) Determine the quadratic function whose zeros are -2 and 5/2

vovangra [49]

So the factors of the quadratic function must be (x+2) and (2x-5) foiling them gives us 2x^2-x-10

3 0

3 years ago

A doctor prescribes a 100-mg antibiotic tablet to be taken every eight hours. Just before each tablet is taken, 20% of the drug

zvonat [6]

Answer:

120 mg

Step-by-step explanation:

20% of 100 is 20 so that would be 20mg plus the 100mg they just took

100+20=120mg

i hope this helps you!

7 0

3 years ago

A business has a savings account that earns a 3% annual interest rate. At the end of 1996, the business had $4,000 in the accoun

olya-2409 [2.1K]

Answer:

The amount the business initially invested in 1991 is approximately;

A. $3,450

Step-by-step explanation:

The question relates to investment and finance

The parameters given are;

The annual interest rate on the business savings account = 3%

The amount the business had in the account at the end of 1996 = $4,000

The given formula used to determine the final amount in the savings account, $F = p \cdot \left(1 + \dfrac{r}{100} \right )^t$

Where;

F = The final amount

p = The initial (principal) investment

r = The annual interest rate

t = The time in years

Therefore, we have;

The required information = The initial amount, 'p', the business invested in 1991

The time in years, 't', from 1991 to 1996 = 1996 - 1991 = 5

∴ The time in years from 1991 to 1996, t = 5 years

The final amount in 1996, F = $4,000

r = 3%

Plugging in the values in the given equation gives;

$4000 = p\cdot \left(1 + \dfrac{3}{100} \right )^5$

Therefore;

$p = \dfrac{4000}{\left(1 + \dfrac{3}{100} \right )^5} = 3,450.43513754$

To the nearest dollar, the initial amount the business invested in 1991, p ≈ $3,450.

3 0

3 years ago

Does anyone know how to solve this math equation?

garik1379 [7]

Answer:

x = 8; y = -6

Step-by-step explanation:

We have a system of equations.

$\dfrac{3(x - 2)}{2} - \dfrac{y - 2}{4} = 11$

$\dfrac{2(x + 2)}{5} - \dfrac{y}{3} = 6$

First, let's multiply both sides of each equation by the LCM of both denominators of that equation to eliminate all denominators.

First equation:

$8 \times \dfrac{3(x - 2)}{2} - 8 \times \dfrac{y - 2}{4} = 8 \times 11$

$12(x - 2) - 2(y - 2) = 88$

$12x - 24 - 2y + 4 = 88$

$12x - 2y = 108$

$6x - y = 54$

The equation above is the simplified first equation.

Second equation:

$\dfrac{2(x + 2)}{5} - \dfrac{y}{3} = 6$

$15 \times \dfrac{2(x + 2)}{5} - 15 \times \dfrac{y}{3} = 15 \times 6$

$6(x + 2) - 5y = 90$

$6x + 12 - 5y = 90$

$6x - 5y = 78$

The equation above is the simplified second equation.

The simplified system of equations is:

$6x - y = 54$

$6x - 5y = 78$

The x term of both equations is 6x. If we multiply both sides of the second equation by -1 to get a first term of -6x, then when it is added to 6x, the x terms are eliminated.

Now we rewrite the first simplified equation as it is. Below it, we write the second equation multiplied by -1 on both sides. Then we add the equations.

$6x - y = 54$

$-6x + 5y = -78$