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3 years ago

2/5 + 3/8 - Adding fractions

Mathematics

2 answers:

m_a_m_a [10]3 years ago

5 0

Answer:31/40

Step-by-step explanation:

To add you have to have a common denominator. To get that you can multiply each fraction by one in different ways. 2/5 *8/8= 16/40 and 3/8* 5/5= 15/40 and then add those two to give you 31/40.

shutvik [7]3 years ago

3 0

Answer:

31/40 is your answer hope this helps you

You might be interested in

1. 2x+5(6x-9)=179<br>2. -40=6x-5(-4x-18)<br>solve for X

pav-90 [236]

Answer:

1. x = 7

2. x = -5

Step-by-step explanation:

1. Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation: 2*x+5*(6*x-9)-(179)=0

Pull out like factors 6x - 9 = 3 • (2x - 3)

(2x + 15 • (2x - 3)) - 179 = 0

Pull out like factors: 32x - 224 = 32 • (x - 7)

Solve: x-7 = 0

Add 7 to both sides of the equation = 7

2. Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation: -40-(6*x-5*(-4*x-18))=0

Pull out like factors: -4x - 18 = -2 • (2x + 9)

-40 - (6x - -10 • (2x + 9)) = 0

Pull out like factors: -26x - 130 = -26 • (x + 5)

-26 • (x + 5) = 0

Solve: x+5 = 0

Subtract 5 from both sides of the equation: x = -5

Hope this helps! Please mark as brainliest.

Thanks!

5 0

2 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

3 years ago

Solve the system of equations y = 2x + 3 and 4x - 2y = -6 using a graphical method.

emmasim [6.3K]

Answer:

-1

Step-by-step explanation:

You are asked to solve this set by substitution. Note that the first equation defines

the value of y in terms of x. Therefore, you can use the right side of this equation

(2x + 3) as a substitution for y in the second equation. When you substitute 2x +3 for y

in the second equation, the result is:

2x + 3 = 4x + 7

Let's set the goal of getting all the terms that contain x isolated on the left side of

the equal sign and all the constants by themselves on the right side. Begin by getting rid

of the +3 on the left side by subtracting 3 from the left side. But if you subtract

3 from the left side you must also subtract 3 from the right side. These subtractions

result in the following sequence:

2x + 3 - 3 = 4x + 7 - 3

Combining numbers on the left side and on the right side simplifies the equation to:

2x = 4x + 4

Now, in a similar fashion, let's get rid of the 4x on the right side by subtracting

4x from the right side. When we do this subtraction, to keep the equation in balance

we must also subtract 4x from the left side. This results in:

2x - 4x = 4x - 4x + 4

On both sides of the equation combine terms containing x to get:

-2x = 4

Finally, solve for x by dividing both sides of the equation by -2 because -2 is the multiplier

of x. This results in:

x = 4/-2 = -2

So now that we know x = -2, we can return to either one of the original equations

in the equation set and substitute -2 for x. Then we can solve for y. Let's return

to the first equation and substitute -2 for x. This begins with:

y = 2x + 3

Then substituting -2 for x gives:

y = 2*(-2) + 3 = -4 + 3 = -1

7 0

3 years ago

What is the zero slope

Mars2501 [29]

It's the first one because all the y's are the same number. That means that when you graph the chart you would get a horizontal line. Horizontal lines on graphs always have a slope of zero....unlike a verticals slope is undefined.

7 0

3 years ago

TUV = 9x + 1 TUW = 7x-9 WUV 5x-11

bazaltina [42]

9x + 1 + 7x - 9 = 5x - 11
16x - 8 = 5x - 11
11x - 8 = -11
11x = -3
-3/11 = x

6 0

3 years ago

Read 2 more answers