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Ivan
2 years ago
14

Solve the equation or inequality 3/2t-16=4/3t-6

Mathematics
1 answer:
Tamiku [17]2 years ago
7 0

Answer:

<em>t=60</em>

Step-by-step explanation:

3/2t-16=4/3t-6

(subtract 4/3t from both sides)

3/2t-16-4/3t=-6

(add 16 to both sides)

3/2t-4/3t=-6+16

(simplify)

1/6t=10

(divide by 1/6 (or multiply by 6 on both sides)

t=60

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A square is constructed on side AD of quadrilateral ABCD such that FA lies on AB, as shown in the figure.
Elza [17]

Answer:  The co-ordinates of F are (4.4, -4.6) and the co-ordinates of D are (4.4, -1.4).

Step-by-step explanation:  Given that a square is constructed on side AD of quadrilateral ABCD such that FA lies on AB as shown in the figure. The co-ordinates of A are (6, -3) and the co-ordinates of B are (10, 1).

Also, AD : AB = 2 : 5 and the co-ordinates of the point E are (2.8, -3).

We are to select the correct co-ordinates of the points F and D.

Let, (a, b) are the co-ordinates of F and (c, d) are the co-ordinates of D.

Since ADEF is a square, so we have

AD = DE = EF = FA.

Given that

AD : AB = 2 : 5, so  FA : AB = 2 : 5.

That is, \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{2.8+6}{2},\dfrac{-3-3}{2}\right)

We have, after applying the internal division formula that

\left(\dfrac{2\times 10+5\times a}{2+5},\dfrac{2\times 1+5\times b}{2+5}\right)=(6,-3)\\\\\\\Rightarrow \left(\dfrac{20+5a}{7},\dfrac{2+5b}{7}\right)=(6,-3)\\\\\\\Rightarrow \dfrac{20+5a}{7}=6,~~~~~\dfrac{2+5b}{7}=-3\\\\\\\Rightarrow 20+5a=42,~~~~\Rightarrow 2+5b=-21\\\\\\\Rightarrow 5a=22,~~~~~~~~~~\Rightarrow 5b=-23\\\\\\\Rightarrow a=4.4,~~~~~~~~~~~\Rightarrow b=-4.6.

So, the co-ordinates of F are (4.4, -4.6).

Now, since ADEF is a square, and the diagonals of a square bisect each other.

So, the mid-points of both the diagonals are same.

That is,

\textup{mid-point of DF}=\textup{mid-point of AE}\\\\\\\Rightarrow \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{2.8+6}{2},\dfrac{-3-3}{2}\right)\\\\\\\Rightarrow \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{8.8}{2},\dfrac{-6}{2}\right)\\\\\\\Rightarrow \dfrac{c+4.4}{2}=\dfrac{8.8}{2},~~~~~~\dfrac{d-4.6}{2}=-\dfrac{6}{2}\\\\\\\Rightarrow c+4.4=8.8,~~~~~\Rightarrow d-4.6=-6\\\\\Rightarrow c=4.4,~~~~~~~~~~~~\Rightarrow d=-1.4.

So, the co-ordinates of D are (4.4, -1.4).

Thus, the co-ordinates of F are (4.4, -4.6) and the co-ordinates of D are (4.4, -1.4).

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2 years ago
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5/6 and 0.833
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Triangle A B C is reflected across line m to form triangle A prime B prime C prime. Triangle A prime B prime C prime is rotated
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Answer:

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Step-by-step explanation:

The problem statement tells you exactly what the transformations are.

The first transformation is reflection across line m.

The second transformation is rotation about point A'.

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These are both rigid transformations, so ΔABC ≅ A'B''C''.

6 0
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