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3 years ago

Sam made a square sign with an area of 410 square inches.What is the approximate perimeter of the square sign?

2 answers:

6 0

Each side of the square = square root of 410 = 20.248 inches

so the perimeter = 4 * 20.248 = 81.7 inches to nearest tenth of an inch

Masteriza [31]3 years ago

4 0

Are you takeing the exam to.

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Find the different between compound interest payable anually and half anually for a amount Rs14000 at the rate of 12•\•per annum

Ostrovityanka [42]

Answer:

$113.08

Step-by-step explanation:

The formula for calculating future value:

FV = P (1 + r/m)^mn

FV = Future value

P = Present value

R = interest rate

N = number of years

m = number of compounding

14,000(1.12)^2 = 17,561.60

14,000 ( 1 + 0.12/2)^(2x2) = 17674.68

Difference : 17674.68 - 17,561.60 = $113.08

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2 years ago

What is the vertical asymptote for this function? y=1/x-2

nataly862011 [7]

Answer: x = 0

Step-by-step explanation: so x = 0 is the equation of the vertical asymptote, and 0 must be left out of the domain: (−∞,0)U(0,∞) in interval notation.

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3 years ago

(24x3 − 14x2 + 20x + 6) ÷ (4x2 − 3x + 5) = Q + R 4x2 − 3x + 5

riadik2000 [5.3K]

Answer:

Q= 6x+1

R= -7x+1

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please please please help me with this...

dlinn [17]

Answer:

It's option d.

Step-by-step explanation:

The line y = x + 2 has domain x < 2 (because of the clear circle.)

The lines y = x + 1 has domain x ≥ 2. ( because of the filled circle).

7 0

3 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

3 years ago