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3 years ago

6

Determine the strongest classification for quadrilateral ABCD with vertices at A(1, 2), B(3, 5), C(5, 3), and D(2, 1). The quadr

ilateral is a because .

1 answer:

Vlada [557]3 years ago

8 0

This is all i could come up with

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An artist wants to frame a square painting with an area of 400 square inches . she wants

Artemon [7]

(a) $x^{2}$ = 400
(b) $\sqrt{x^2} = \sqrt{400}$
$x=20,-20$
(c) Only the positive solution makes sense because you cannot have a negative side length.
(d) Since one side of the painting is 20 inches, all four sides together will be 20*4 = 80 inches

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3 years ago

How many solutions does the following equation have ?

bija089 [108]

There is only one solution

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3 years ago

Read 2 more answers

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi

Margaret [11]

Answer:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=26Kg$

Step-by-step explanation:

Data

Brine a (Ba)

$V_{Ba}=5\frac{Lt}{min}\\ Concentration(Bca)=0.05\frac{Kg}{Lt}$

Brine b (Bb)

$V_{Bb}=10\frac{Lt}{min}\\ Concentration(Bcb)=0.04\frac{Kg}{Lt}$

we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)

, then the amount of salt (y) left in the tank after (t) minutes: $y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}$

Finally:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=0.65\frac{Kg}{min}(40min)=26Kg$

being y(t) the amount of salt (y) per unit of time (t)

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2 years ago

The sum of two numbers is 12 and their difference is 6. What are the two numbers?

wolverine [178]

Answer:

The two numbers are 9 and 3.

Step-by-step explanation:

x+y=12

x-y=6

---------

2x=18

x=18/2

x=9

9-y=6

y=9-6

y=3

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3 years ago

Hi please help me it’s due in 45 min

Rina8888 [55]

Answer:

what is the result of part D

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3 years ago