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lesya [120]
3 years ago
7

this is a grade 4 maths question. i need help with doing a model from this question as well. thank you! —————————————————- a rop

e was cut into 2 pieces. The first piece was twice the length of the second piece. If the first piece was 5m 50cm long what was the length of the rope before it was cut
Mathematics
1 answer:
NeX [460]3 years ago
4 0

Answer:825cm

Step-by-step explanation:550cm/2=275cm

275*3=825cm

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baherus [9]

Step 1

Given;

4x^2-y^2+24x+4y+28=0

Required; To find the center that eliminates the linear terms

Step 2

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Step 3

Substitute a,d,e into the vertex form

\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\  \\  \end{gathered}

Step 4

Completing the square for -y²+4y

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Step 5

Substitute a,d,e into the vertex form

\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}

Step 6

\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}

Step 7

\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}

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Step-by-step explanation:

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3 years ago
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SOLUTION:

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A simple sketch will help;

We can use the sine rule to solve for the angle B;

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Thus;

m\angle B=67.34^o

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