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3 years ago

10

-2 1/2 - (-1 3/4) ?

1 answer:

Olin [163]3 years ago

8 0

Answer:

- 3/4

Step-by-step explanation:

-2 1/2 - (-1 3/4)

Subtracting a negative is like adding

-2 1/2 + (1 3/4)

Get a common denominator of 4

-2 2/4 + 1 3/4

The signs are different so we subtract and take the sign of the larger

2 2/4 - 1 3/4

Borrowing from the 2

1 4/4 + 2/4 - 1 3/4

1 6/4 - 1 3/4

3/4

2 2/4 was larger and it was negative so we add a negative sign

- 3/4

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21% of the students at a High are in Maths group, 15 % are in science group, and 6 % are in both. If a student is selected rando

mihalych1998 [28]

Answer:

6%

Step-by-step explanation:

The probability of picking a maths student = 21% = 21/100

The probability of picking a science student = 15% = 15/100 = 3/20

The probability of picking in both = 6/100 = 3/50

The probability that the student is in maths or science = 21/100 - 15/100 = 6/100 = 6%.

4 0

3 years ago

A client is instructed to take

nasty-shy [4]

It would be 3/2 x 3/1 which would = 4 1/2

3 0

3 years ago

Read 2 more answers

A 1080 foot cable weighs 307 pounds. How much will 36 inches of cable weigh? Round to 2 decimal places

noname [10]

Answer:

0.85 pounds

Step-by-step explanation:

1080 foot cable = 307 pounds

36 inches/2.9999988ft = (2.9999988 x 307) / 1080 = 920.9996316/1080 = 0.85277743666 pounds ≈ 0.85 pounds

1 inch = 0.0833333 ft

36 inch = 36 x 0.0833333 = 2.9999988 ft

6 0

4 years ago

2. Multiply 6⁄5 × 25⁄24 .

andrew-mc [135]

6*25=150 and 5*24=120 so now we have a fraction of uh 150/120 :P Then we have to simplify and divide both numbers by 30 so 150 divided by 30 is 5 and 120 divided by 30 is 4. So now we have a fraction of 5/4. If you need it as a mixed number, then im pretty sure the answer is 11/4 :D

5 0

3 years ago

y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the

vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

$y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},$

where $c_1, c_2 \in \mathbb{R}$ . We now want to find a solution $y$ such that $y(-1)=3$ and $y'(-1)=-3$ . Therefore, all we need to do is find the constants $c_1$ and $c_2$ that satisfy the initial conditions. For the first condition, we have: $y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.$

For the second condition, we need to find the derivative $y'$ first. In this case, we have:

$y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.$

Therefore:

$y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.$

This means that we must solve the following system of equations:

$\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.$

If we add the equations above, we get:

$\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e} \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.$

If we now substitute $c_1 = 0$ into either of the equations in the system, we get:

$c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}$

This means that the solution obeying the initial conditions is:

$\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.$

Indeed, we can see that:

$y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3$

$y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,$

which do correspond to the desired initial conditions.

3 0

3 years ago