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3 years ago

60+6=48 What is the value of b?

Mathematics

1 answer:

mina [271]3 years ago

3 0

Answer:

-2

Step-by-step explanation:

Assuming you mean 60 + 6b = 48

subtract 60 from both sides: 6b = 48 - 60 = -12

Now divide both sides by 6 to solve for b: b = -12/6 = -2

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AB=45. If RT=3(x+9), what is the value of x?

Margaret [11]

6 because 45 divides by 3 equals 15 and fifteen minus 9 equals 6, if you plug 6 into the equation it makes sense

7 0

3 years ago

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What is the solution to y=5x+1 and y=-5x+15?

Semenov [28]

Answer:

x=7/5, y=8

Step-by-step explanation:

You can use substitution.

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3 years ago

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#7. What are the values/ how do I do it

Tasya [4]

Answer:

x =2

SK = 21

KY = 13

SY = 34

Step-by-step explanation:

we can see from the picture the following:

SK + KY = SY equation 1

SK=13x-5 equation 2

KY = 2x+9 equation 3

SY = 36-x equation 4

using equation 2, equation 3 and equation 4 in equation 1 we have:

13x - 5 + 2x + 9 = 36-x

15x +4 = 36 - x

15x +x = 36-4

16x = 32

x = 32/16

x= 2

using x as 2 we evaluate it in equation 2, equation 3 and equation 4:

SK=13(2)-5 = 21

KY= 2(2) +9 = 13

SY = 36-2 = 34

5 0

3 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

4 years ago

Select any 3 side lengths that make an obtuse triangle.

faltersainse [42]

That is obtuse it doesn’t have equal sides

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3 years ago