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k0ka [10]
2 years ago
8

Please help with 2. 3. and 4.

Mathematics
2 answers:
Y_Kistochka [10]2 years ago
6 0

Answer:

2. Since it is equilateral, we just set up an equation with using any two sides (since they are all equal)

12x - 22 = 10x - 6

12x - 10x - 22 = -6

2x - 22 = -6

2x = 22 - 6

2x = 16

x = 16/2

x = 8

Each side = 74

3. Same formula for this. Get any two sides and setup the equation.

4x - 25 = x + 14

4x - x - 25 = 14

3x = 25 + 14

3x = 39

x = 39/3

x = 13

Each side = 27

4. This one is also simple. Since DE = EF. We substitute its sides with the equations.

8x - 13 = 5x + 17

8x - 5x - 13 = 17

3x - 13 = 17

3x = 17 + 13

3x = 30

x = 30/3

x = 10

DE and EF = 67

DF = 31

Mashcka [7]2 years ago
5 0

Answer:

Step-by-step explanation:

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Solve for [x]. The polygons in each<br> 2x-20<br> 16<br> 16<br> 8
Nataly_w [17]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:x = 16

____________________________________

\large \tt Solution  \: :

If two polygons are similar, their corresponding sides ratios will be equal to each other.

\qquad \tt \rightarrow \:  \cfrac{8}{16}  =  \cfrac{2x - 20}{24}

\qquad \tt \rightarrow \:  \cfrac{1}{2}  =  \cfrac{2x - 20}{24}

\qquad \tt \rightarrow \: 2x - 20 =  \cfrac{24}{2}

\qquad \tt \rightarrow \: 2x = 12 + 20

\qquad \tt \rightarrow \: 2x = 32

\qquad \tt \rightarrow \: x = 16

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

7 0
1 year ago
What is the ratio of 24 to 32
lubasha [3.4K]
It is 3 to 4
you find the GCF of 24 and 32 which is 8. then divide 24 and 32 by 8 to find the ratio
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3 years ago
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THIS IS URGENT. The table above shows the number of students that prefer a particular type of pizza. If there are 25 students
german

Answer:

40%

Step-by-step explanation:

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In triangle QRS, QR = 8 and RS = 5. Which expresses all possible length of side QS?
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Obviously it's answer D) Why?

a) any side of a triangle is SMALLER than the SUM of the 2 others

b) any side of a triangle is GREATER than the DIFFERENCE of the 2 others

Hence D) 8-5 < QS < 8+5==>3 < QS < 13
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Some body can help me with a geometric mean maze
Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.

2. South-East point from the Start: By the 2nd theorem,

x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.

3. West point from the previous: By the 2nd theorem,

x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.

4. West point from the previous: By the 1st theorem,

9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.

5. West point from the previous: By the 2nd theorem,

10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.

6. North point from the previous: By the 1st theorem,

x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.

7. East point from the previous: By the 2nd theorem,

x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.

8. North point from the previous: By the 1st theorem,

x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.

8. West point from the previous: By the 2nd theorem,

x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.

9. North point from the previous: By the 1st theorem,

12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.

101. East point from the previous: By the 1st theorem,

6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.

11. East point from the previous: By the 2nd theorem,

20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.

12. South-east point from the previous: By the 2nd theorem,

18^2=x\cdot 21.6\Rightarrow x=15.

13. North point=The end.

6 0
3 years ago
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