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2 years ago

13

Which of the following inequalities would have a dotted boundary line?

1 answer:

aliya0001 [1]2 years ago

5 0

D because y is only 'greater than' all the others could also be 'equal to'

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Seven more then 3 rimes a number is 49

NikAS [45]

7 <span> more than 3 times a number </span>
<span>more than is code for + </span>
<span>times is multiply </span>
<span>so 7+3n=31 </span>

<span>solve: </span>
<span>3n=24 </span>
<span>n= 8</span>

6 0

3 years ago

Read 2 more answers

Rodrigo bought a pumpkin pie for $4.85 and a can of whipped cream for $1.55. He gave the cashier $7.00. How much change should R

madam [21]

Answer:
$0.60

step-by-step explanation:
4.85 + 1.55 = 6.40
7.00 - 6.40 = 0.60

7 0

2 years ago

The price of a sweater was reduced from $20 to $12 by what percentage was the price of the sweater reduced

CaHeK987 [17]

It should be a 40% reduction :)

5 0

2 years ago

The price P of a good and the quality Q of a good are linked.

Irina-Kira [14]

the equilibrium point, is when Demand = Supply, namely, when the amount of "Q"uantity demanded by customers is the same as the Quantity supplied by vendors.

That occurs when both of these equations are equal to each other.

let's do away with the denominators, by multiplying both sides by the LCD of all fractions, in this case, 12.

$\bf \stackrel{\textit{Supply}}{-\cfrac{3}{4}Q+35}~~=~~\stackrel{\textit{Demand}}{\cfrac{2}{3}Q+1}\implies \stackrel{\textit{multiplying by 12}}{12\left( -\cfrac{3}{4}Q+35 \right)=12\left( \cfrac{2}{3}Q+1 \right)} \\\\\\ -9Q+420=8Q+12\implies 408=17Q\implies \cfrac{408}{17}=Q\implies \boxed{24=Q} \\\\\\ \stackrel{\textit{using the found Q in the Demand equation}}{P=\cfrac{2}{3}(24)+1}\implies P=16+1\implies \boxed{P=17} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{Equilibrium}{(24,17)}~\hfill$

3 0

2 years ago

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

3 years ago