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slamgirl [31]
3 years ago
6

True or False: X=3 is the solution to 2x+10/5=4. 1. True 2. False

Mathematics
2 answers:
s2008m [1.1K]3 years ago
8 0

Answer:

Step-by-step explanation:

False

maksim [4K]3 years ago
6 0
2x + 10/5 = 4

If the *divided by 5* is for the whole expression
2x + 10 = 20
2x = 20 - 10
2x = 10
x = 10/2 = 5 [false]

If *divided by 5* is only for 10
2x + (10/5) = 4
2x + 2 = 4
2x = 4-2
x = 2/2
x = 1 [false]

Answer will be false in both cases but i gave the solution to make you understand.
Hope it helps.
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The mean monthly rent of students at Oxnard University is $890 with a standard deviation of $206. John's rent is $1,395. What is
mezya [45]

Answer:

$299

Step-by-step explanation:

Rent+standard deviation 890+206= $1,096

John's rent: $1,395

Z-score: 1395 - 1096=$299

8 0
4 years ago
Read 2 more answers
Find the sum 2 2/3 + 2 2/3
Len [333]

Answer: 5 and 1/3

Step-by-step explanation: To add mixed numbers, first add the fractions.

So here, we have 2/3 + 2/3 which is 4/3.

Then add the whole numbers.

So we have 2 + 2 which is 4.

So our answer is 4 and 4/3.

But notice, that since 4/3 is an improper fraction, or answer is not in lowest terms. 4/3 can be rewritten as the mixed number 1 and 1/3. So 4 and 4/3 is the same as 4 + 1 and 1/3 which is 5 and 1/3.

Since 5 and 1/3 is in lowest terms, this is our final answer.

So we can say that 2 and 2/3 + 2 and 2/3 is 5 and 1/3

5 0
3 years ago
01.01)<br> Evaluate 32 + (8 − 2) ⋅ 4 − 6 over 3. (1 point)
kicyunya [14]

Answer:

In the explanation. :)

Step-by-step explanation:

Find the exact value using trigonometric identities.

50 , 3  p o i n t

Hope this helps. Have a great day!

6 0
4 years ago
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
Sean buys 5 packages of fish. there's 7/8 pounds of fish in each package. what is the total weight in pounds of fish that sean b
ehidna [41]
5packages*7/8pounds=35/8 or 4 3/8 or 4.375 pound of fish

Answer=4 3/8 pounds of fish
4 0
4 years ago
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