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andrew-mc [135]
3 years ago
14

The diagram shows a floor in the shape of a trapezium. Tim is going to paint the floor. Each 5 litre tin of paint costs £16.99 a

nd 1 litre of paint covers an area of 1.9 m2 How much would it cost to buy all the paint he needs? You must show how you get your answer. 0 m 16m £16.99
Mathematics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer: £186.89

Step-by-step explanation:

-We are given that each 5l  tin of paint costs £16.99 . We therefore divide the paint used by 5 and multiply by £16.99:

Hence, it will cost £186.89 to buy all the paint

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3/10 is represented as 30%.
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What would be the slope of the positions (3,3) and (4,5)
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Answer:

m = 2

Step-by-step explanation:

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I will give brainliest quick
a_sh-v [17]

There are 44 total small egg cartons, 264 / 6 = 44

There are 176 total medium egg cartons, (264 / 3) x 2 = 176

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3 0
2 years ago
Help me with question a please ! With full workings !
frosja888 [35]
A)


\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;Q&({{ 0}}\quad ,&{{ 2}})\quad &#10;%  (c,d)&#10;P&({{ 0.5}}\quad ,&{{ 0}})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2}&#10;\\\\\\&#10;QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4}&#10;\\\\\\&#10;QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)


c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;Q&({{ 0}}\quad ,&{{ 2}})\quad &#10;%   (c,d)&#10;R&({{ -0.5}}\quad ,&{{ 0}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}

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