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3 years ago

(8v + 8) (8v^2 + 4v - 8) find the product

Mathematics

2 answers:

mars1129 [50]3 years ago

5 0

Answer:

Answer is below

Step-by-step explanation:

64v^3+96v^2-32v-64

Nikolay [14]3 years ago

5 0

Answer:

-64

Step-by-step explanation:

(8×v+8)×((8×v^2+4×v)−8)

(0+8)×((8×v^2+4×v)−8)

8×((8×v^2+4×v)−8)

8×((8×0+4×v)−8)

8×((0+4×v)−8)

8×((0+0)−8)

8×(0−8)

8×-8

-64

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please help me, Prove a quadrilateral with vertices G(1,-1), H(5,1), I(4,3) and J(0,1) is a rectangle using the parallelogram me

mestny [16]

Answer:

Step-by-step explanation:

We are given the coordinates of a quadrilateral that is G(1,-1), H(5,1), I(4,3) and J(0,1).

Now, before proving that this quadrilateral is a rectangle, we will prove that it is a parallelogram. For this, we will prove that the mid points of the diagonals of the quadrilateral are equal, thus

Join JH and GI such that they form the diagonals of the quadrilateral.Now,

JH= $\sqrt{(5-0)^{2}+(1-1)^{2}}=5$ and

GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$

Now, mid point of JH= $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

= $(\frac{5+0}{2},\frac{1+1}{2})=(\frac{5}{2},1)$

Mid point of GI= $(\frac{5}{2},1)$

Since, mid point point of JH and GI are equal, thus GHIJ is a parallelogram.

Now, to prove that it is a rectangle, it is sufficient to prove that it has a right angle by using the Pythagoras theorem.

Thus, From ΔGIJ, we have

$(GI)^{2}=(IJ)^{2}+(JG)^{2}$ (1)

Now, JI= $\sqrt{(4-0)^{2}+(3-1)^{2}}=\sqrt{20}$ and GJ= $\sqrt{(0-1)^{2}+(1+1)^{2}}=\sqrt{5}$

Substituting these values in (1), we get

$5^{2}=(\sqrt{20})^{2}+(\sqrt{5})^{2} }$

$25=20+5$

$25=25$

Thus, GIJ is a right angles triangle.

Hence, GHIJ is a rectangle.

Also, The diagonals GI= $\sqrt{(4-1)^{2}+(3+1)^{2}}=5$ and HJ= $\sqrt{(0-5)^2+(1-1)^2}=5$ are equal, thus, GHIJ is a rectangle.

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