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2 years ago

Find the number of real solutions of each equation using the discriminant.

Mathematics

1 answer:

lianna [129]2 years ago

8 0

Answer: 1: x^2 + 25 = 0 x=5

2: x^2 - 11x + 28 = 0 x=7 i think

3: x^2 + 8x + 16 = 0 x=-4

Step-by-step explanation:

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The Flynns' monthly budget allowed them to spend 5555 for household expenses. During June and July they

cricket20 [7]

Answer:

They have over spent $925

Step-by-step explanation:

Given:

The Monthly amount Flynn's spent on Household = $555

Amount spent In June and July together = #2035

To Find:

The over spent amount in June and July = ?

Solution:

For a month the budget is $555

According to the budget the amount they should have spent for the month of June and July will be

=> $555 for June + $555 for July

=> $1110

But they have spent $ 2035 for June and July together

Si the extra amount will be

Amount spent - actual budget amount

=> $2035 - $1110

=> $925

3 0

2 years ago

Let log base aU=X and log base aV=Y, then a to the x power =? and a to y power =?

Andrews [41]

Answer:

${ \bf{ log_{a}(U) = x}} \\ { \boxed{ \tt{ {a}^{x} = U}}} \\ \\ { \bf{ log_{a}(V) = y}} \\ { \boxed{ \tt{ {a}^{y} = V }}}$

3 0

2 years ago

Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on

boyakko [2]

Answer:

Mathew invested $600 and $2400 in each account.

Solution:

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest $(I_1)$ earned in first account for one year,

$\text {simple interest}=\frac{\text {pnr}}{100}$

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get $(I_1)$ as,

$I_{1}=\frac{P \times 1 \times 3}{100}$ ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest $(I_2)$ earned in second account,

$I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2$

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

$I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3$

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

$I = I_1+ I_2 ---- eqn 4$

By substituting eqn 1 , 2, 3 in eqn 4

$\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}$

$\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}$

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600

Hence, Mathew invested $600 and $2400 in each account.

3 0

2 years ago

Read 2 more answers

Please help me with question 24 ‍♀️

love history [14]

for answer 2 is A=1/32

for answer 1 is=A=4.8

hope that helps you

3 0

2 years ago

The price of Veronica’s meal before tax and tip was $11.92. Veronica paid 8% tax, then added a 15% tip to the total. To the near

Alecsey [184]

Answer:

$14.80

Step-by-step explanation:

The price of Veronica's meal before tax and tip was $11.92. Veronica paid 8% tax, then added a 15% tip to the total

8 0

2 years ago