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2 years ago

10

I NEED HELP WITH MATH QUESTION PLEASE HELP ME NO LINKS !!!

2 answers:

lukranit [14]2 years ago

6 0

Answer:

21.8

Step-by-step explanation:

1.9 + 9 = 10.9

10.9 x 2 = 21.8

Margarita [4]2 years ago

3 0

Answer:

21.8

Step-by-step explanation:

2×1.9=3.8

2×9=18

total=21.8

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Which expressions are equivalent to ? Check all that apply.

GrogVix [38]

Answer:

6.2+8.7+8.9x

Step-by-step explanation:

For expressions to be equivalent, they must simplify to the same lowest form. 8.9x+6.2+8.7 simplifies to 8.9x +14.9.

Each of the expressions below must simplify to 8.9x + 14.9 to be equivalent.

9x+6+9 = 9x + 15 NO

3.9+6.2+8.7x = 8.7x + 10.1 NO

3.7+8.9+6.2 = 18.8 NO

6.2+8.7+8.9 = 23.8 NO

6.2+8.7+8.9x = 8.9x + 14.9 YES

3.9+6.2x+8.7 = 6.2x + 12.6 NO

3.9x+8.7+6.2 = 3.9x + 14.9 NO

8 0

3 years ago

Read 2 more answers

A publisher sells 10⁶ copies of a new book. Each book sells 10² pages. How many pages total are there in all of the books sold?

KatRina [158]

Answer:

10^12.

Step-by-step explanation:

You need to solve 10^6 * 10^2 without using standard notation. As with variables, if you had x^6 * x^2 you would add the exponents. Replace x with 10. Your answer is 10^8. As a sidenote, only multiply exponents when raising a power to another power, e.g. (10^6)^2 = 10^12.

3 0

3 years ago

Read 2 more answers

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

What was the affect of frequent fighting in the Middle East during the carters administration?

astraxan [27]

B. innocent lives were lost

7 0

3 years ago

Read 2 more answers

Carrie has 170 coins. She has 10 times as mary coins as she had last month.

asambeis [7]

Answer:

17coins

Step-by-step explanation:

cuz 10 × y = 170

170 ÷ 10 = y

17 = y

3 0

3 years ago