Help me with this pls anyone

Mathematics

1 answer:

podryga [215]3 years ago

5 0

Answer:

Step-by-step explanation:

x^2=3x-2(subtract 3x and add 2) x^2-3x+2=0(group) (x-1)(x-2)=0(find x) x=1;x=2

find y values for x values and check:

1^2=1, pair 1:(1,1); 3(1)-2=3-2=1, same pair

2^2=4, pair 2:(2,4); 3(2)-2=6-2=4, same pair

pairs: (1,1);(2,4)

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W • (-4+ z) = mz + 17

Setler [38]

Answer:

z = (17+4w)/(w-m)

Step-by-step explanation:

w • (-4+ z) = mz + 17

Distribute

-4w +wz = mz+17

Subtract mz from each side

-4w +wz - mz = mz+17-mz

-4w +wz-mz = 17

Add 4w to each side

-4w +4w+wz-mz = 17+4w

wz-mz = 17+4w

Factor out z

z(w-m) = 17+4w

Divide by (w-m)

z(w-m)/(w-m) = (17+4w)/(w-m)

z = (17+4w)/(w-m)

3 0

3 years ago

If two triangles have equal perimeters, then they could also have which of the following? Select all that apply.

Radda [10]

Answer: The answers is (B) equal areas.

Step-by-step explanation: Given that two triangles have equal perimeters.

As shown in the attached figure, let us consider two right-angles triangles, ΔABC and ΔDEF, with sides AB = 3 cm, BC = 4 cm, AC = 5 cm, DE = 4 cm, EF = 3 cm and DF = 5 cm.

So the perimeters of both the triangles = 3 + 4 + 5 = 4 + 3 + 5 = 12 cm.

Since volume term is not valid in case of triangles, so they cannot have equal volumes. Therefore, option (A) is incorrect.

Area of ΔABC is

$A_{ABC}=\dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times 3\times 4=6~\textup{cm}^2,$

and area of ΔDEF is

$A_{DEF}=\dfrac{1}{2}\times DE\times EF=\dfrac{1}{2}\times 4\times 3=6~\textup{cm}^2.$

Therefore, they may have equal areas and so option (B) is correct.

If the triangles have equal bases, then the heights will also be equal and both the triangles will be same. Similar is the case with equal heights. So, options (C) and (D) are incorrect.

Thus, the correct option is (B). equal areas.