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\(\Rightarrow a=\sqrt{100}=10\left(cm\right)\)

\(b=\sqrt{36}=6\left(cm\right)\)

Pythagoras \(\Rightarrow\)\(c=\sqrt{10^2-6^2}=8\left(cm\right)\)

\(\Rightarrow\)\(S_C=8^2=64\left(cm^2\right)\)

Condittion: \(3x+1\ne0\)

\(\Rightarrow x\ne-\dfrac{1}{3}\)

\(\dfrac{x-1}{3x+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x-1\right)=3x+1\)

\(\Leftrightarrow2x-2=3x+1\)

\(\Leftrightarrow2x-3x=1+2\)

\(\Leftrightarrow-x=3\)

\(\Leftrightarrow x=-3\) (satisfy)

\(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}+\dfrac{3-\sqrt{5}}{3+\sqrt{5}}\)

\(=\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}+\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}\)

\(=\dfrac{\left(3+\sqrt{5}+3-\sqrt{5}\right)\left(3+\sqrt{5}-3+\sqrt{5}\right)}{4}\)

\(=\dfrac{6\cdot2\sqrt{5}}{4}=\dfrac{12\sqrt{5}}{4}=3\sqrt{5}\)

\(\left(100+1\right)\cdot100:2=5050\)

Answer: A

Because \(\Delta ABC\) is a square triangle.

\(\Rightarrow\)\(AB^2+AC^2=BC^2\) (Pytago)

\(\Rightarrow BC=\sqrt{3^2+4^2}=5\)

We have:
\(AB\cdot AC=BC\cdot AH\)

\(\Rightarrow AH=\dfrac{AB\cdot AC}{BC}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}=2.4\left(cm\right)\)

Answer: 2.4 cm

\(\Delta PQS\) has: \(PQ=PS\)

\(\Rightarrow\)​\(\Delta PQS\) is an ​ isosceles triangle.

\(\Rightarrow\)\(\widehat{PSQ}=\frac{180^0-\widehat{QPS}}{2}\)

\(\Rightarrow\)\(\widehat{PSQ}=84^0\)

\(\Rightarrow\)\(\widehat{PSR}=180^0-84^0=96^0\)

\(\Delta PSR\) has: \(SP=SR\)

\(\Rightarrow\)\(\Delta PSR\) is an isosceles triangle

\(\Rightarrow\)\(\widehat{SPR}=\frac{180^0-\widehat{PSR}}{2}\)

\(\Rightarrow\)\(\widehat{SPR}=42^0\)

\(\Rightarrow\)\(\widehat{QPR}=12^0+42^0=54^0\)

Answer: C.

\(\Delta{AOC}\) have:

\(\widehat{A}+\widehat{C}+\widehat{AOC}=180^0\) (1)

\(\Delta{BOD}\) have:

\(\widehat{B}+\widehat{D}+\widehat{BOD}=180^0\) (2)

\(\widehat{AOC}=\widehat{BOD}\) (vertically opposite angles) (3)

(1)(2)(3)

\(\Rightarrow\)\(\widehat{A}+\widehat{C}=\widehat{B}+\widehat{D}\)

\(3AA1⋮9\)

\(\Leftrightarrow\left(3+1+A+A\right)⋮9\)

\(\Leftrightarrow\left(4+2A\right)⋮9\)

\(\Leftrightarrow A=7\)

According to the problem, we have:

\(\widehat{A} = 3 \widehat{B}\)\(\Rightarrow\)\(\widehat{B}=\frac{\widehat{A}}{3}\)  (1)

\(\widehat{C}=2\widehat{A}\) (2)

\(\widehat{A} + \widehat{B} + \widehat{C}=180^0\) (3)

Change (1)(2) to (3) we have:

\(\widehat{A} + \frac{\widehat{A}}{3}+2\widehat{A}=180^0\)

\(\frac{10}{3}\widehat{A}=180^0\)

\(\widehat{A} = 54^0\)

\(\Rightarrow\) (C) is correct

Have: \(2\cdot2+3\cdot4+4\cdot3=28\left(wheels\right)\)in the garage

a) We have:

\(3\left(2x-1\right)^2\ge0\forall x\)

\(7\left(3y+5\right)^2\ge0\forall y\)

\(\Rightarrow\left\{{}\begin{matrix}3\left(2x-1\right)^2=0\\7\left(3y+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(3y+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-1=0\\3y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

Answer: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(x^2+y^2-2x+10y+26=0\)

\(\left(x^2-2x+1\right)+\left(y^2+10y+25\right)=0\)

\(\left(x-1\right)^2+\left(y+5\right)^2=0\)

We have:

\(\left(x-1\right)^2\ge0\forall x\)

\(\left(y+5\right)^2\ge0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-5\end{matrix}\right.\)

Answer: \(\left\{{}\begin{matrix}x=1\\y=-5\end{matrix}\right.\)

To cut a wooden rod to 35 pieces, we must make 34 cuts.

According to the problem, it takes 3 minutes to make 1 cut

So: \(3\cdot34=102\left(minutes\right)\) to complete the cuts

a) 46;64

b) We have:
\(3=0+1+2\)

\(6=1+2+3\)

\(11=2+3+6\)

\(\Rightarrow6+11+20=37\)

Because:

+) Kate is the 18^th from the front:

\(\Rightarrow\)Kate is the 18^th child (from the front)

+) John is the 32^nd from the back:
​​\(\Rightarrow\)John is the 8^th child (from the front)

\(\Rightarrow\)There are 9 children between Kate and John

There are 6 ways she can arrange them:

Way 1: red, pink, yellow

Way 2: red, yellow, pink

Way 3: pink, red, yellow

Way 4: pink, yellow, red

Way 5: yellow, red, pink

Way 6: yellow, pink, red

Mai's stamps: \(40+7=47\left(stamps\right)\)

Answer: Mai has 47 stamps

Mai's notebooks: \(4+3=7\left(notebooks\right)\)

James's notebooks: \(4+7=11\left(notebooks\right)\)

Answer: James has 11 notebooks

Mai's stamps: 40 + 7 = 47 (stamps)

The total height of 2 boys: \(131.6\cdot2=263.2\left(cm\right)\)

The total height of 4 girls: \(128.3\cdot4=513.2\left(cm\right)\)

The total height of all children: \(263.2+513.2=776.4\left(cm\right)\)

All children: \(2+4=6\left(children\right)\)

The average height of all children: \(\dfrac{776.4}{6}=129.4\left(cm\right)\)

The legs of 4 horses: \(4\cdot4=16\left(legs\right)\)

The legs of  11 people: \(11\cdot2=22\left(legs\right)\)

The total legs: \(16+22=38\left(legs\right)\)