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3 years ago

9

I need the answer and dont put a random wrong answer just because you need points i really need to know this answer

2 answers:

aksik [14]3 years ago

8 0

You are right bro that what I always get

olganol [36]3 years ago

6 0

Answer:

16

Step-by-step explanation:

The answer is 16 its hard to explain

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PLEASE HELP im super confused

FinnZ [79.3K]

Answer:

7 Quarters and 3 dimes

Step-by-step explanation:

You have to find how many quarters (0.25) and dimes (0.10) are in $2.05. There are 7 quarters, which will add up to $1.75. You can't have 8 quarters because then you would have 2 dollars, and there wouldn't be enough for a dime, there would only by 5 cents left, or a nickle. Then, you find out how many nickels you need, which is 3.

.10x3=.30. 1.75+0.3=2.05.

I hope you understood:)

4 0

3 years ago

Read 2 more answers

Solve for f: d/2+f/4=6

soldi70 [24.7K]

I believe this should be right

3 0

3 years ago

X2-7x+12byx-3 divide of algebraic expression<br><br>

Leokris [45]

-41.

Explanation and show work:

5 0

3 years ago

Read 2 more answers

Which is a critical point of the function f(x) = sin(2x) in the interval [0,2pi] ?

MrRissso [65]

Answer:

D pi/4, 3pi/4

Step-by-step explanation:

Differentiate the function then equal it to zero and solve for the interval (see attached workings)

5 0

3 years ago

Find the solution of the given initial value problems in explicit form. Determine the interval where the solutions are defined.

natali 33 [55]

Answer:

The solution of the given initial value problems in explicit form is $y=x-x^2-2$ and the solutions are defined for all real numbers.

Step-by-step explanation:

The given differential equation is

$y'=1-2x$

It can be written as

$\frac{dy}{dx}=1-2x$

Use variable separable method to solve this differential equation.

$dy=(1-2x)dx$

Integrate both the sides.

$\int dy=\int (1-2x)dx$

$y=x-2(\frac{x^2}{2})+C$ $[\because \int x^n=\frac{x^{n+1}}{n+1}]$

$y=x-x^2+C$ ... (1)

It is given that y(1) = -2. Substitute x=1 and y=-2 to find the value of C.

$-2=1-(1)^2+C$

$-2=1-1+C$

$-2=C$

The value of C is -2. Substitute C=-2 in equation (1).

$y=x-x^2-2$

Therefore the solution of the given initial value problems in explicit form is $y=x-x^2-2$ .

The solution is quadratic function, so it is defined for all real values.

Therefore the solutions are defined for all real numbers.

4 0

3 years ago