All categories

2 years ago

A pair of jeans is marked down at 25% off the original price. The original price of

Mathematics

1 answer:

ozzi2 years ago

5 0

Answer:

$51.38

Step-by-step explanation:

1 - .25 = .75

you're paying 75% of the price

68.50 • .75 = 51.375 ≈ $51.38

You might be interested in

Solve for x. Round to the nearest tenth.

Misha Larkins [42]

Answer:

Step-by-step explanation:

Youre given the adjacent and hypotenuse, so you can use cosine. SOHCAHTOA

cos(37) = 48/x

x = 60.1

5 0

2 years ago

7. Given the function f(x)= x - 3x +4, evaluate f(5)-|(2).

SVETLANKA909090 [29]

Yes that’s correct i did this before

3 0

3 years ago

I dont know how to answer this question

sweet [91]

Answer:

So, where is the question that you are talking about?

8 0

3 years ago

I want to just get in with life.

Ahat [919]

Answer:

The equation would be y = -2x + 1

Step-by-step explanation:

All you have to do is put it in the form y = mx + b where m is the slope and b is the y-intercept (for example, the equation y= 2x + 1 has a slope of 2 and a y-intercept of 1.)

Since you have slope, or m, = -2 and the y-intercept, or b, = 1, the equation would be y = -2x + 1

4 0

2 years ago

Read 2 more answers

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

3 years ago