Ask question

Ask question

All categories

3 years ago

6

Natalie buys organic almonds priced at $77 from the grocery store. How much did she pay the cashier if she received $23 in chang

e?

1 answer:

alexandr402 [8]3 years ago

3 0

She paid $100 because she bought the almonds for $77 then she received $23 back (100-77=23) (77+23=100)

You might be interested in

7 times a number is 18 less than the square of that number. Find the positive solution.

Arte-miy333 [17]

So 7 x = x2 -18 so you get the quadratic x^2 -7x -18 and then get 2 x -9= -18 and 2 + -9 = -7 so you get (x + 2) ( x - 9) so the positive solution would be x-9 = 0 so x= 9. Hope this helps :)

7 0

3 years ago

**GIVING OUT BRAINLIEST PLS HELP**<br> Which inequality represents the graph below?

BartSMP [9]

Answer:

r≥2.5

it can't be < because the green dot is filled in

r is larger than 2.5 because the arrow goes to the right

7 0

3 years ago

Read 2 more answers

Sin(-x)= -cos x for all values of x. True or false

kogti [31]

Answer:

That's incorrect. The simplest way to show this is by evaluating the functions at a given point. Let's say x=0, then:

Sin(-x) = Sin(0) = 0

-cos x = -cos (0) = -1

Therefore, Sin(-x)≠-cos x.

5 0

4 years ago

Read 2 more answers

Ms Happy has decided to start wearing a face mask, because it is now required of her to do so. The mask she has started wearing

Alchen [17]

Answer:

37.20 %

Step-by-step explanation:

Circumference = 2 × pi × r

30 = 2 × 3.14 × r

r = 4.7770700637

Area of her face = pi × r²

3.14 × 4.7770700637²

= 71.6560509556 in²

Area of the mask = 9 × 5 = 45 in²

Uncovered area

= 71.6560509556 - 45

= 26.6560509556 units²

% of uncovered area:

26.6560509556/71.6560509556 × 100

37.2%

6 0

4 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

4 years ago