Parallelogram ABCD is a rectangle.

Mathematics

2 answers:

yanalaym [24]2 years ago

5 0

Answer:

$y=3$

Step-by-step explanation:

Given: Parallelogram ABCD is a rectangle.

$AC=5y-2$ and $BD=4y+1$

To find: Value of <em>y</em>

Solution: It is given that ABCD is a rectangle, and we know that the opposite sides of a rectangle, and the diagonals of a rectangle are always equal.

As, AC and BD are the diagonals of the rectangle, both will be equal.

So, on equating AC and BD we have,

$AC=BD$

$5y-2=4y+1$

$5y-4y=1+2$

$y=3$

Hence, the value of <em>y </em>is 3.

weeeeeb [17]2 years ago

3 0

Because diagonals of a parallelogram are equal, set both problems equal to one another. Solving, you should get y=3

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HELP!!!In the expression 100 a/b, if a is greater than b, which statements are true? b ≠ 0. Select all that apply.

MAXImum [283]

Answer:

$\large\boxed{\text{The expression is equal to}\ \sqrt[b]{100^a}}\\\boxed{\text{The expression is greater than 100}}$

Step-by-step explanation:

$a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\text{True statement:}\\\\100^\frac{a}{b}=\sqrt[b]{100^a}\\\\a>b\to\dfrac{a}{b}>1\to\dfrac{a}{b}=1\dfrac{a-b}{b}\\\\100^\frac{a}{b}=100^{1\frac{a-b}{b}}=100^{1+\frac{a-b}{b}}=100^1\cdot100^\frac{a-b}{b}=100\sqrt[b]{100^{a-b}}\\\\a>b\to a-b>0\to100^{a-b}>1\\\\\text{Therefore}\ 100\sqrt[b]{100^{a-b}}>100$