All categories

4 years ago

HELP PLEASE!

Mathematics

2 answers:

dalvyx [7]4 years ago

4 0

Answer: The correct options are:

2: 50 ≤ s ≤ 100

5: She could deposit $50

6: She could deposit $75

hope this helps :)

Archy [21]4 years ago

3 0

Answer:

Option 2 50 ≤ s ≤ 100

Option 5 She could deposit $50

Option 6 She could deposit $75

Step-by-step explanation:

Let

s -----> amount of money Layla deposit into a saving account

we know that

25%=25/100=0.25

50%=50/100=0.50

$s\geq 0.25*200$ -----> $s\geq \$50$

$s\leq 0.50*200$ -----> $s\leq \$100$

The compound inequality is

$\$50 \leq s\leq \$100$

Verify each case

case 1) 25 ≤ s ≤ 50

The statement is false

see the procedure

case 2) 50 ≤ s ≤ 100

The statement is True

see the procedure

case 3) s ≤ 25 or s ≥ 50

The statement is false

Because is s ≤ 100 and s ≥ 50

case 4) s ≤ 50 or s ≥ 100

The statement is false

Because is s ≤ 100 and s ≥ 50

case 5) She could deposit $50

The statement is true

Because the value of s satisfy the compound inequality $\$50 \leq s\leq \$100$

case 6) She could deposit $75

The statement is true

Because the value of s satisfy the compound inequality $\$50 \leq s\leq \$100$

You might be interested in

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$