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3 years ago

What is the midpoint of (-3, 4) and (-6, -1)

Mathematics

1 answer:

pishuonlain [190]3 years ago

6 0

Do the steps of x and y. And see where they hit each other

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Solve. Show your work. 0.24+ 0.532 =

garik1379 [7]

Answer:

0.772

Step-by-step explanation:

0.24+ 0.532 =

Line up the decimals

0.24

+ 0.532

------------

0.772

7 0

3 years ago

Read 2 more answers

-5z + 9 =24 solve the equation

nataly862011 [7]

Answer:

z=-3

Step-by-step explanation:

-5z+9=24 so, subtract 9 from both sides becuase of liked terms to get -5z=15, then divide both sides by -5 to get z = -3 hope it helps:)

6 0

3 years ago

C=Wtc/1000 for W please I need help it's algebra

DedPeter [7]

C1000÷ct=w

you pretty much ×1000 and then ÷t and c

3 0

3 years ago

We were never taught this in class, and I understand it, kind of... but i'm just confused with everything.

GaryK [48]

$\bf \cfrac{x^3y}{xy^5}\cdot \cfrac{x^2y^9}{x^8}\implies \cfrac{x^3x^2y^1y^9}{x^1x^8y^5}
\\ \quad \\
\textit{now, using exponent rules, same base, different exponent}
\\ \quad \\
\cfrac{x^{3+2}y^{1+9}}{x^{1+8}y^5}\implies \cfrac{x^5y^{10}}{x^9y^5}\\
----------------------------\\$ $\bf \textit{now, keep in mind that}
\\ \quad \\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad

\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
% negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 

\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}
\\ \quad \\
thus\\
----------------------------\\
$ $\bf \cfrac{x^5y^{10}}{x^9y^5}\implies \cfrac{x^5y^{10}}{1}\cdot \cfrac{1}{x^9}\cdot \cfrac{1}{y^5}\implies x^5y^{10}\cdot x^{\boxed{-9}}\cdot y^{\boxed{-5}}
\\ \quad \\
x^5x^{-9}y^{10}y^{-5}\implies x^{5-9}y^{10-5}\implies x^{-4}y^5
\\ \quad \\
\cfrac{1}{x^{\boxed{4}}}\cdot \cfrac{y^5}{1}\implies \cfrac{y^5}{x^4}$

8 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Maru [420]

Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$

6 0

3 years ago