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3 years ago

Estimate the number of pints in 445 fluid ounces?

Mathematics

1 answer:

Dovator [93]3 years ago

5 0

27.812 us liquid pints

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If the simple interest on $6,000 for 9 years is $3,240 , then what is the interest rate?

igor_vitrenko [27]

Rate of interest is 6 %

Solution:

Given in question that,

simple interest = $ 3240

number of years = 9 years

principal sum = $ 6000

To find: interest rate

The simple interest is given as:

$S.I = \frac{pnr}{100}$

Where, "p" is the principal

"r" is the rate of interest

"n" is the number of years

Substituting the values in above formula,

$3240 = \frac{6000 \times 9 \times r}{100}\\\\3240 = 60 \times 9 \times r\\\\3240 = 540 \times r\\\\r = \frac{3240}{540}\\\\r = 6$

Thus rate of interest is 6 %

8 0

3 years ago

Need help just with number 2 please and thank you!!

BaLLatris [955]

I believe the answer is 17

4 0

3 years ago

NO LINKS

Elza [17]

So the answer is 30 packages but I did 25 times 20 = 500 and then 25 times 10 = 250
So. 30 = 750

6 0

3 years ago

Read 2 more answers

How do you illustrate quadratic equation in one varible?

kvasek [131]

Answer:

A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable. One absolute rule is that the first constant "a" cannot be a zero.

Step-by-step explanation:

6 0

3 years ago

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago