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Vaselesa [24]
3 years ago
7

Solve the following inequality.

Mathematics
1 answer:
user100 [1]3 years ago
5 0

Answer:

A

Step-by-step explanation:

Given

21 ≤ - 3(x - 4) < 30 ← divide all 3 intervals by - 3

Remembering to reverse the direction of the signs as a result of dividing by a negative quantity

- 7 ≥ x - 4 > - 10 ← add 4 to each interval

- 3 ≥ x > - 6, that is

- 6 < x ≤ - 3 → A

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Maslowich
Ok start off by finding the base first and the equation of finding the area of the square is A=bh
A=11 x 11= 121 cm^2

Next find the area of one triangle and the equation is A=1/2bh
A=1/2(11)(10)
A=55 cm^2

Multiply the area of one triangle by 4 since there are four triangles and it should equal to 220 cm^2. You then add this to the base and the final answer should be 341 cm^2.

I hope this helps
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3 years ago
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Daniel says that more than half the students spent 2 1/2 hours or more on homework.
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Answer: Because

Step-by-step explanation: Yeah im good

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What is the simplest form of 72/75?
GaryK [48]
I hope this helps you



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Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,
rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t ,  y_2(t) = 0   \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1 $ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t  \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0,  ⇒ dx = 0

y = 9 t,  ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$ \oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx  $

                        = 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy $

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

                            $= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$  is conservative.

5 0
3 years ago
123 Savings and Loan charges a monthly fee of $8 on checking accounts and an overdraft protection fee of $33. Neela's check regi
Nimfa-mama [501]

Answer:

The account statement had an ending balance of $153

Step-by-step explanation:

Check register actual balance= $256

Less; Written check= -$312

The account is short; $256-$312= -$56

Add the new deposit =-$56+$250= $194

Therefore , new account balance= $194

Subtract fees($8 & $33) from the balance= $194- $8- $33=$153

The account statement had an ending balance of $153

4 0
3 years ago
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