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lawyer [7]
3 years ago
10

Prove that PQ + QR + RS + SP = 0 in the given quadrilateralPQRS​

Mathematics
1 answer:
Katarina [22]3 years ago
7 0

Step-by-step explanation:

Using ∆ law of vector addition,

PR→=PQ→+QR→PR→=PQ→+QR→ - (i)

Also,

RP→=RS→+SP→RP→=RS→+SP→ - (ii)

Taking LHS,

=PQ→+QR→+RS→+SP→=PQ→+QR→+RS→+SP→

[Put value of PR→=PQ→+QR→PR→=PQ→+QR→ from equation (i) ]

=(PR→)+RS→+SP→=(PR→)+RS→+SP→

[Put value of RP→=RS→+SP→RP→=RS→+SP→ from equation (ii) ]

=PR→+RP→=PR→+RP→

=PR→−PR→=PR→−PR→

=0=0

RHS

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Bob and James are finishing the roof of a house. Working alone, Bob can shingle the roof in 14 hours. James can shingle the same
Snowcat [4.5K]

we know that Bob can do the whole job in 14 hours, how much of the work has he done in 1 hour only?  well since he can do the whole lot in 14 hours in 1 hour he has only done 1/14 th of the job.

we know that James can do it in 18 hours, a bit slower, so in 1 hour he has done only 1/18 th of the job.

let's say it takes both of them working together say "t" hours, so in 1 hour Bob has done (1/14) of the work whilst James has done (1/18) of the work, the whole work being t/t or 1 whole, so for just one hour that'd 1/t done by both Bob and James.

\bf \stackrel{Bob}{\cfrac{1}{14}}~~+~~\stackrel{James}{\cfrac{1}{18}}~~=~~\stackrel{total~for~1~hour}{\cfrac{1}{t}} \\\\\\ \stackrel{\textit{using an LCD of 126}}{\cfrac{9+7}{126}=\cfrac{1}{t}}\implies \cfrac{16}{126}=\cfrac{1}{t}\implies 16t=126\implies t=\cfrac{126}{16} \\\\\\ \stackrel{\textit{7 hrs, 52 minutes and 30 seconds}}{t=\cfrac{63}{8}\implies t=7\frac{7}{8}}\implies \stackrel{\textit{rounded up}}{t=7.88}

7 0
3 years ago
A rectangular plot of ground is 30 meters longer than it is wide. its area is 10,000 square meters.
dimulka [17.4K]
A = W· L
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Width of a plot is 86.12 m and lenght is 116.12 m.
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3 years ago
Can any experts help me??<br> Solve for x° and y°.<br> Easy for anyone.
Annette [7]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
Can pretty please help me with 2 math question its due today
Mashcka [7]
Answer:
(4x + 12 ) - ( -8 ) =
4x + 20
Explanation:
Just add them up and form a equation ( this is for the second one )
5 0
3 years ago
Read 2 more answers
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