Ask question

Ask question

All categories

2 years ago

5

The ratio of white chocolate to dark chocolate sold at a candy shop was 4 : 3. If there were 20 bars of white chocolate sold, ho

w many bars of dark chocolate were sold?

1 answer:

Ksju [112]2 years ago

5 0

Answer:

15

Step-by-step explanation:

if the white chocolate is the 4 you would need to multiply it by 5 to get 20 so if u multiply 3 by 5 you get 15 hope this helps :)

You might be interested in

I need to calculate a=8c-b divided by c, how do I calculate?

laila [671]

Answer:

Step-by-step explanation:

8xC-Bdivide by c

5 0

3 years ago

12% of people are wearing sunglasses. If

VashaNatasha [74]

Answer:

500 people?Not too sure but that is my best guess

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Andrew pays a monthly membership fee of $10 at his gym. Each time he uses the gym, he pays $5. Last month, Andrew spent a total

Ivenika [448]

Answer:

11

Step-by-step explanation:

If we input 11 into the equation, we get: 10 + 5(11) = 65.

5 x 11 = 55

55 + 10 = 65

Therefore x = 11

6 0

2 years ago

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

2 years ago

Evaluate 50 + 0.5 ×(41 - 32)

Zarrin [17]

To solve this expression, we need to follow the order of operations (BEDMAS)

50+0.5(41-32)
=50+0.5(9)
=50+4.5
=54.5

Your answer is 54.5

7 0

2 years ago