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Margarita [4]
3 years ago
8

There is a gap of 1 metre you put a table into gap.The table is 70 centimetres wide.How much space is left?

Mathematics
2 answers:
USPshnik [31]3 years ago
7 0
100cm=1m
70cm= x
You cross multiply
100x=70
Divide both sides by 100
.:x=7/10 m or 0.7m
1m-0.7m=0.3m
Fofino [41]3 years ago
4 0
.3 meter or 30 cemimeters
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What is the solution to the system of linear equations represented by the matrix below?
olga_2 [115]
\begin{bmatrix}2&4\\1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}6\\3\end{bmatrix}

Notice that the second row is half the first, so the rows are dependent. This means there are infinitely many solutions to the system.
6 0
3 years ago
Read 2 more answers
Sorry for asking all these questions, this is just really confusing
Fittoniya [83]

Answer:

-31

Step-by-step explanation:

hope this helped! <3

6 0
3 years ago
PLEASE HELP ASAP! URGENT!
Nonamiya [84]

Answer:

It would be a negative number, no matter what negative integer you put it as.

Step-by-step explanation:

For example, lets substitute -2

f(-2) = -8-4-1 = -13; a negative number

4 0
3 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
Please answer my question bcz I can’t understand
RUDIKE [14]
Please send more context as of the "green numbers".

For the multiplication,

i 49 x 10 = 490
490 ÷ 10 = 49

ii 2.3 <span>÷ 10 = 0.23
0.23 x 10 = 2.3

iii 0.034 x 1000 = 34
34 </span><span>÷ 1000 = 0.034

iv 876 </span><span>÷ 100 = 8.76
8.76 x 100 = 876

Hope this helps :)</span>
3 0
3 years ago
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