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2 years ago

Given: Rhombus ABCD. If mBAD is 64 degrees, then the mDCA is ___ degrees.

Mathematics

1 answer:

Marta_Voda [28]2 years ago

6 0

Answer: 32 degrees

Step-by-step explanation:

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Чу- 8х + 5x - Зу<br> OOoO

natulia [17]

Y-3x

Explanation: do 4y-3y which equals 1 which just equals Y. Then do -8x+5x which equals -3x

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3 years ago

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What is 3 times minus 5 plus 23 times minus 9

Verizon [17]

answer:

-1

step-by-step explanation:

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3 years ago

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4. The length of a rectangle is one more than twice the width. The area is 105 in?.

Zanzabum

9514 1404 393

Answer:

7 in

Step-by-step explanation:

For width w in inches, the length is given as 2w+1. The area is the product of length and width, so we have ...

A = LW

105 = (2w +1)w

2w^2 +w -105 = 0

To factor this, we're looking for factors of -210 that have a difference of 1.

-210 = -1(210) = -2(105) = -3(70) = -5(42) = -6(35) = -7(30) = -10(21) = -14(15)

So, the factorization is ...

(2w +15)(w -7) = 0

Solutions are values of w that make the factors zero:

w = -15/2, +7 . . . . . negative dimensions are irrelevant

The width of the rectangle is 7 inches.

7 0

3 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

3 years ago

Tripp and Rico are two dogs. Tripp weighs exactly 35 pounds more than Rico. Together, they weigh exactly 49 pounds. How much doe

babymother [125]

Call the weights of the two dogs $t$ and $r$ .

We know that Tripp weights 35lbs more than Rico, or:
$t=r+35$

We also know that the total weight of both dogs is 49lbs.:
$t+r=49$

Now, by substitution:
$(r+35)+r=49$
$2r+35=49$
$2r+35-35=49-35$
$2r=14$
$r=\frac{14}{2}$
$r=7$
Rico weighs 7lbs.

Then:
$t=r+35$
$t=7+35$
$t=42$
Tripp weighs 42lbs.

To check our work:
$t+r=49$
$42+7=49$
$49=49$ CHECK!

5 0

3 years ago