The length of this market on the map is:

    128,5 x 1:5000=0,0257(m)=2,57 (cm)

 Answer:2,57 cm

The number of students participate in one of three sports is:

             78-4-5-2-3=64(students)

The number of students who play only baseball is:

           64-7-1=56(students)

The least amount he should pay for the bill is:

        19,50 x 80%=15,6(dollars)

The least amount he should pay for the tax is:

        15,6 x 6%=0,936(dollars)

The least amount he should pay fỏr the tip is:

        19,50 x 20%=3,9(dollars)

The longest distance between any two vertices is:

           2:1 x 3=6(ft)=6 x 0,3048=1,8288\(\approx\)1,83(m)

Answer:1,83 m

Because (-4,-5) and (4, 2) is a value equivalent two consecutive positive integers is bigger than 2 and smaller than 5.Thence inferred:Two integers is 3 and 4

So the sum of those two integers is:3+4=7

Chúng ta có: 13 =2+11=3+10=5+8=7+6=13+0=3 +3 +7

So the number need find is:3 x 3 x 7=63

Answer:63

We have diagrams:

Barbara's cell phone bill :|____|____|____|____|____|____|____|

Tina's cell phone bill      :|____|____|____|____|____|

Equal parts of Barbara's cell phone bill minus equal parts Tina's cell phone bill is:

            7-5=2(equal parts)

Barbara's bill is:

             14:2 x 7=49

We have:\(2^{18}=4^x\Leftrightarrow2^{18}-4^x=0\Leftrightarrow\left(2^2\right)^9-4^x=0\Leftrightarrow4^9-4^x=0\)

\(\Leftrightarrow4^z\left(4^{9-x}-1\right)=0\).Because 4^x>0 so 4^9-x=1\(\Rightarrow9-x=0\Rightarrow x=9\)

Apply theorem into the triangle between three square labeled L,J and K ,we have the area of the square labeled J is:

7²-2²=49-4=45 cm²

So the area of the square labeled J is 45 cm²

wrong topic

Answer:

Common fraction:\(\dfrac{12+24+44+52}{68+12+24+44+52}=\dfrac{132}{200}=\dfrac{33}{50}\)

Answer:

Common fraction:\(\dfrac{1}{25}\)

We have: | x-1 |+|x-3|+|x-5|+|x-7|=|x-1|+|x-3|+|5-x|+|7-x|

\(\ge\)|x-1+x-3+5-x+7-x|=|8|=8

Mark ''='' occurs when \(\left(x-1\right)\left(x-3\right)\left(5-x\right)\left(7-x\right)\ge0\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)\ge0\)

Case 1 :x-1 , x-3 , x-5 , x-7 \(\ge0\) so x=7 because if x>7 so (x-1)(x-3)(x-5)(x-7)>0(not satisfy)

Case 2 :x-1 , x-3\(\ge\)0 ; x-5 , x-7\(\le0\)\(\Rightarrow x-3\ge0;x-5\le0\)\(\Rightarrow5\ge x\ge3\)

Case 3 :x-1 , x-5 \(\ge\)0 ; x-3 , x-7\(\le\)0\(\Rightarrow x-5\ge0;x-3\le0\)\(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\)(not satisfy)

Case 4 :x-1 , x-7 \(\ge0\) ; x-3 , x-5\(\le\)0\(\Rightarrow\left\{{}\begin{matrix}x\ge7\\x\le3\end{matrix}\right.\)(not satisfy)

Case 5 :x-1 , x-3 , x-5 , x-7\(\le\)0\(\Rightarrow x\le1\)

Case 6 :x-1 , x-3\(\le0\) ; x-5 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

Case 7 :x-1 , x-5\(\le\)0 ; x-3 , x-7\(\ge\)0\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge7\end{matrix}\right.\)(not satisfy)

Case 8 :x-1 , x-7\(\le0\) ; x-3 , x-5\(\ge0\)\(\Rightarrow\left\{{}\begin{matrix}x\le1\\x\ge5\end{matrix}\right.\)(not satisfy)

Retry x=7 we have |7-1|+|7-3|+|7-5|+|7-7|=12(not satisfy)

So max_x=5

Put a,b sequence is the number of toy animals Jerry and Ben have:

Deduced:the legs of toy animals Jerry has is 4a;the legs of toy animals Ben has is 2b

We have:\(\left\{{}\begin{matrix}a-b=10\\4a+2b=220\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b=10\\2a+b=110\end{matrix}\right.\)\(\Rightarrow3a=120\Leftrightarrow a=40\)

So toy animals Jerry has is 40

Put  the total number of pennies Mollie has is:a

We have\(\left\{{}\begin{matrix}a-3⋮5\\a-1⋮7\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a-8⋮5\\a-8⋮7\end{matrix}\right.\)\(\Rightarrow a-8\in BC\left(5,7\right)=\left\{0,35,70,105,.......\right\}\)

\(\Rightarrow a\in\left\{8,43,78,113,.........\right\}\)

Because a\(⋮\)3 and a<100 so a=78

So the total number of pennies Mollie has is 78 pennies

         40% of 80 is:

           80.40%=32

         32% of 75 is:

           75.32%=24

        40% of 80 minus 32% of 75 is:

         32-24=8

        75% of 8 is:

           75%.8=6

                   Answer:6

We have:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{xy}{12}=\dfrac{x^2}{9}=\dfrac{108}{12}=9\Rightarrow x^2=81\)

\(\Rightarrow x=\pm9\)\(\Rightarrow y=\pm12\)

So \(\left(x,y\right)\in\left\{\left(-9,-12\right);\left(9,12\right)\right\}\) satisfy problem