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Novosadov [1.4K]
3 years ago
7

Simplify by combining like terms. − 5 p + 4 m − 2 p + m

Mathematics
2 answers:
IrinaK [193]3 years ago
7 0

Answer:

-7p  +5m

Step-by-step explanation:

− 5 p + 4 m − 2 p + m

Combine like terms

-5p -2p   +4m +m

-7p  +5m

Ainat [17]3 years ago
6 0

Answer:

-7p+5m

Step-by-step explanation:

See image below:) :)

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Find a unit vector that has the same direction as the given vector. 8i ? j + 4k
e-lub [12.9K]

Answer with Step-by-step explanation:

a.Let given vector

a=8i-j+4k

\mid a\mid=\sqrt{(8)^2+(-1)^2+(4)^2}=9

By using the formula

Magnitude of a vector=\sqrt{x^2+y^2+z^2}

Where x=Coefficient of i

y=Coefficient of j

z=Coefficient of k

Unit vector=\hat{a}=\frac{a}{\mid a\mid}

By using the formula

The unit vector=\hat{a}=\frac{8i-j+4k}{9}=\frac{8}{9}i-\frac{1}{9}j+\frac{4}{9}k

b.Let vector b=-2i+4j+2k

\mid b\mid=\sqrt{(-2)^2+4^2+2^2}=2\sqrt 6

\hat{b}=\frac{-2i+4j+2k}{2\sqrt 6}

Length of vector=6

Therefore, the vector in the direction of <-2,4,2> with length 6 is given by =6\hat{b}=6\times \frac{-2i+4j+2k}{2\sqrt 6}

The vector in the direction of <-2,4,2> with length 6 is given by =-\sqrt 6(-i+2j+k)

c.\theta=\frac{\pi}{3}

\mid v\mid=8

Let v=v_x i+v_y j

v_x=\mid v\mid cos\theta

v_x=8cos\frac{\pi}{3}=8\times \frac{1}{2}=4

cos\frac{\pi}{3}=\frac{1}{2}

v_y=\mid v\mid sin\theta

v_y=8\times sin\frac{\pi}{3}=8\times \frac{\sqrt 3}{2}=4\sqrt 3

sin\frac{\pi}{3}=\frac{\sqrt 3}{2}

Therefore, the vector v in component form=4i+4\sqrt 3j

7 0
3 years ago
HELP ASAP PLEASE thanks youu
BlackZzzverrR [31]

Answer: Probably A

Step-by-step explanation: You need to use a ruler for this, because without one its impossible to get it exact

Using the ruler though, you'd measure each side of the wall and either use a ratio of 0.5in:4ft or multiply every 0.5in by 4ft.

6 0
3 years ago
Please answer this for me
coldgirl [10]

In 1-4, to determine whether a sequence is either arithmetic or geometric, you need to look at differences of consecutive terms (arithmetic) and ratios of consecutive terms (geometric). If you can't find it, the sequence will fall under the "neither" category.

For example, the differences between consecutive terms in the first sequence are

\left\{2-4,\dfrac12-2,\dfrac14-\dfrac12,\ldots\right\}=\left\{-2,-\dfrac32,-\dfrac14,\ldots\right\}

If the sequence was arithmetic, the difference between consecutive terms would have been the same constant throughout this list. But that's not the case, so this sequence is not arithmetic.

The ratios between consecutive terms are

\left\{\dfrac24,\dfrac{\frac12}2,\dfrac{\frac14}{\frac12},\ldots\right\}=\left\{\dfrac12,\dfrac14,\dfrac12,\ldots\right\}

The sequence would have been geometric if the list contained the same value throughout, but it doesn't. So this sequence is neither arithmetic nor geometric.

Meanwhile, in the second sequence, the differences are

\{-1-(-6),4-(-1),9-4,\ldots\}=\{5,5,5,\ldots\}

so this sequence is arithmetic.

In 5-6, you know the sequences are arithmetic, so you know that they follow the recursive rule

a_n=a_{n-1}+d

For example, in the fifth sequence we know the first term is a_1=4. The common difference between terms is d=9-4=5. So using the rule above, we have the pattern

a_2=a_1+d

a_3=a_2+d=a_1+d(2)

a_4=a_3+d=a_1+d(3)

and so on, so that the n-th term is determined entirely by a_1 with the formula

a_n=a_1+d(n-1)

This means the 21st term in the fifth sequence is

a_{21}=a_1+5(21-1)=4+5(20)=104

The process is simple: identify a_1 and d, plug them into the formula above, then evaluate it at whatever n you need to use.

8 0
3 years ago
The volume of a cube is 216mm3.What is the length of its side?
Studentka2010 [4]
Steps:
216 mm3 is the volume.

Given the volume, you can take the cube root of the volume to find the side.
The cube root of 216 is 6.

Answer: 6
8 0
3 years ago
Find the flux of F = x^3 i  + y^3 j  + z^3k through the closed surface bounding the solid region x^2 + y^2 ≤ 4, 0 ≤ z ≤ 4
givi [52]
Use the divergence theorem. Let R be the cylindrical region, then

\displaystyle\iint_{\partial R}\mathbf F\cdot\mathbf n\,\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dV

(where \mathbf n denotes the unit normal vector to \partial R, but we don't need to worry about it now)

We have

\mathrm{div }\mathbf F=(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial\mathbf F}{\partial x}+\dfrac{\partial\mathbf F}{\partial y}+\dfrac{\partial\mathbf F}{\partial z}
\nabla\cdot\mathbf F=3x^2+3y^2+3z^2

For the solid R with boundary \partial R, we can set up the following volume integral in cylindrical coordinates for ease:

\displaystyle3\iiint_R(x^2+y^2+z^2)\,\mathrm dV=3\int_{z=0}^{z=4}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}(r^2+z^2)r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz
=\displaystyle6\pi\int_{z=0}^{z=4}\int_{r=0}^{r=2}(r^3+rz^2)\,\mathrm dr\,\mathrm dz
=\displaystyle12\pi\int_{z=0}^{z=4}(2+z^2)\,\mathrm dz
=352\pi
5 0
3 years ago
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