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3 years ago

BRAINLIESTTT ASAP!!!!!!!!!

Mathematics

1 answer:

Tju [1.3M]3 years ago

6 0

Find where the blue and red lines touch
the answer would be -3.5 and -0.5

the first choice is the correct one

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Kaynard's first step in solving this rational equation is to simplify it by multiplying both sides by a common denominator. Whic

bulgar [2K]

x^2 - 9 = (x + 3)(x -3)

3x - 9 = 3(x - 3)

and

Common denominator would be 6(x+3)(x - 3)

Answer: 6(x+3)(x - 3)

5 0

3 years ago

One-fourth of r subtracted from ten is greater than five

rosijanka [135]

Im not 100% about how this question is explained or written but i think its 8

8 0

3 years ago

Read 2 more answers

If log9 x = 3/2 what is the value of x ? 1 3/2 2 3 27/2 4 . 27

sweet-ann [11.9K]

log 9 X= 3/2

x= 3.5

or x= 7/2

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

3 years ago

There are two spinners. Each spinner has 10 equal sectors labeled with the numbers 1 through 10.

Reika [66]

Answer: Second option is correct.

Step-by-step explanation:

Since we have two spinners,

Each spinner has 10 equal sectors labeled with the numbers from 1 to 10.

Primes numbers from 1 to 10 is given by

$\{2,3,5,7\}$

So, number of outcomes shows a primes number from 1 to 10 = 4

Similarly ,

Composite numbers from 1 to 10 is given by

$\{4,6,8,9,10\}$

So, number of outcomes shows a composite number from 1 to 10 =5

∴ Total outcomes show a prime number on the first spinner and a composite number on the second spinner is given by

$4\times 5=20\\$

Thus, Second option is correct.

3 0

3 years ago