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aleksklad [387]
3 years ago
6

An artist is painting the outside of the box shown below.

Mathematics
1 answer:
AVprozaik [17]3 years ago
3 0

Answer:

15 square feet

Step-by-step explanation:

Multiply 1.5 by 3 which equals 4.5

Multiply 1.5 by 2 which equals 3

Add the above answers (4.5 plus 3)

Then multiply by 2 (7.5 times 2)

It equals 15

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What is the greatest common factor?
dsp73

Answer:

the greatest common factor would be -8w^4x^2

Step-by-step explanation:

lets factor

(-8w^4x^2(x^2w-1)

8 0
3 years ago
Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p
GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2

you replace t1 from the first equation in the second equation:

(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}

Then, for t2 = 0 and t2=1/2 you obtain for t1:

t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths  intersect are:

r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)

B) You have the following two trajectories of two independent particles:

r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t

From the first equation you have:

t=1+2t\\\\t=-1

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0
3 years ago
Gabriella made 36 cookies. She gave away 28 cookies. Use the equation 28 + c = 36 to find c, the number of cookies she kept.
Lera25 [3.4K]

c=8

Hope it helps Hve a nice day

6 0
2 years ago
Read 2 more answers
Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).
ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

\overrightarrow {PQ}==

\overrightarrow {PR}==

The nonzero vector orthogonal to the plane through the points P,Q, and R is

\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}

\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}

Expand along row 1.

\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}

\overrightarrow n=i(-9)-j(8)+k(2)

\overrightarrow n=-9i-8j+2k

\overrightarrow n=

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

8 0
3 years ago
FAST
sertanlavr [38]

Answer: The system consists of parallel lines

Step-by-step explanation:

Given system of lines :

y=\dfrac13x-4

3y-x=-7

Substitute y=\dfrac13x-4 in 3y-x=-7,  we get

3(\dfrac13x-4)-x=-7\\\\\Rightarrow\ x-12-x=-7\\\\\Rightarrow\ -12=-7 , which is not true.

That means , system has no solution.

i.e. they are representing parallel lines. [Parallel lines do not intersect and hence they do not have solution.]

6 0
2 years ago
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