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3 years ago

If stanley bought a sandwich, that cost $3 with 50% off. what was the original price?

Mathematics

2 answers:

Andrew [12]3 years ago

3 0

Answer:

The answer is 6$. Half of 6$ is 3$

STALIN [3.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

Let the original price be =Y,

50% of the original = Y×50/100 = 0.5Y

∴ 50% off =Y - 0.5Y = 3

0.5Y=3

divide both side by the co-efficient 0f Y i.e 0.5

⇒0.5Y/0.5=3/0.5

⇒Y= $6

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Michael invested 5,000 in an account that has a 5.5% annual interest rate. What equation best describes investment after T years

ivolga24 [154]

Answer:

V = 5000 + 275*T for simple annual interest

or: A = 5000 * (1.055)^T for an annual compound interest

Step-by-step explanation:

I assume this is a simple interest rate. If not I will give the one for compound interest.

V = 5000 + 5000* 0.055 * T (Value of account after T years)

V = 5000 + 275*T for simple annual interest

or: A = 5000 * (1.055)^T for an annual compound interest

5 0

3 years ago

Can someone pls. help me with this, the instruction says "Find the Unknown length marked on the following figure. Round all the

coldgirl [10]

Answer:

The length of diagonal d is 14.1421 cm

Step-by-step explanation:

We are given square

Length of side of square = 10 cm

We need to find the length of diagonal d

To find diagonal of square, the formula used is:

$Diagonal\: of\: square=s\sqrt{2}$

where s is length of side of square.

Putting values of s and finding length of diagonal of square

$Diagonal\: of\: square=s\sqrt{2}\\Diagonal\: of\: square=10\sqrt{2}\\Diagonal\: of\: square=10(1.41421)\\Diagonal\: of\: square=14.1421\:cm$

So, The length of diagonal d is 14.1421 cm

3 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$