Can someone please help me with this? It would be greatly appreciated:) thanks

Mathematics

1 answer:

kati45 [8]3 years ago

3 0

Answer:

d < 1

Step-by-step explanation:

First, we have to get d on the same side of the equation. So we will subtract d from both sides and the equation will be 12d-d < + 11. We could also write this problem as d(12-1) < 11, which when simplified would be d(11) < 11. Next, divide both sides by 11 and you get d< 1. Hope this helps and have a great day!!:D

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E%7B2%7D-1%7D%7B2-5a%7D%20times%20%5Cfrac%7B15a-6%7D%7Ba%5E%7B2%7D%2B5a-6%7D" id=

stepan [7]

$\bf \cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\\\\
-----------------------------\\\\
recall\quad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
thus\quad a^2-1\iff a^2-1^2\implies (a-1)(a+1)
\\\\\\
now\quad a^2+5a-6\implies (a+6)(a-1)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{a^2-1}{2-5a}\times \cfrac{15-6}{a^2+5a-6}\implies \cfrac{(a-1)(a+1)}{2-5a}\times \cfrac{3(5a-2)}{(a+6)(a-1)}\\\\
-----------------------------\\\\
$

$\bf now\quad 3(5a-2) \iff -3(2-5a)\\\\
-----------------------------\\\\
thus
\\\\\\
\cfrac{\underline{(a-1)}(a+1)}{\underline{2-5a}}\times \cfrac{-3\underline{(2-5a)}}{(a+6)\underline{(a-1)}}\implies \cfrac{-3(a+1)}{a+6}$

6 0

3 years ago

A school is designing two rectangular parking lots. The second parking lot is being designed so that its perimeter is 3/4 of per

Nataly [62]

The perimeter of second plot is 180 feet

<em><u>Solution:</u></em>

Given that,

The second parking lot is being designed so that its perimeter is 3/4 of perimeter of the first parking lot

<h3><u>Find the perimeter of first plot:</u></h3>

Perimeter = 2(length + width)

From given figure in question,

length = 40 feet

width = 80 feet

Therefore,

Perimeter = 2(40 + 80)

Perimeter = 2(120) = 240

Thus, we got,

Perimeter of first parking plot = 240 feet

Also, given that,

$\text{Perimeter of second plot }= \frac{3}{4} \times \text{Perimeter of first parking plot}\\\\\text{Perimeter of second plot }= \frac{3}{4} \times 240\\\\\text{Perimeter of second plot }= 180$