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Lubov Fominskaja [6]
3 years ago
10

HELP ME PLZZZZZZZZZZ

Mathematics
1 answer:
kogti [31]3 years ago
7 0

Answer:

In both cases, we have similar figures.

This means that the shape of the figures is the same, but the size is different:

PQST is similar to STNR and to NRPQ

VUYZ is similar to YZWX and to VUWX

this means that, for example, in problem 18, the ratio between ST and NR must be the same as the ratio between PQ and ST. This happens because the measure increases by the same scale factor.

With this in mind, we can solve the problem:

18)

ST = 7.5

NR = 5.5

Then the quotient ST/NR is:

ST/NR = 7.5/5.5

And, as we said above:

PQ/ST = ST/NR

PQ/7.5 = 7.5/5.5

PQ = (7.5/5.5)*7.5 = 10.23

19) Here we should have:

YZ/VU = WX/YZ

Then:

22.9/35 = WX/22.9

(22.9/35)*22.9 = WX = 14.98

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Given System of equation:

x-y =6                                  .....,[1]

2x-3z = 16                           ......[2]  

2y+z = 4                              .......[3]

Rewrite the equation [1] as

y = x - 6                                .......[4]

Substitute the value of [4] in [3], we get

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then, we have

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Add 12 to both sides of an equation:

2x-12+z+12=4+12

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On substituting equation [2] in [5] we get;

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Add 3z both sides of an equation:

z+3z = -3z+3z

4z = 0

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z = 0

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2x = 16

Divide by 2 both sides of an equation:

\frac{2x}{2} =\frac{16}{2}

Simplify:

x= 8

Substitute the value of x =8 in equation [4] to solve for y;

y = 8-6 = 2

or

y = 2

Therefore, the solution for the given system of equation is;  x = 8 , y = 2 and z =0

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