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oee [108]
3 years ago
10

A trapezoid has two bases that measure 11 cm and 8 cm. The height of the figure is 5 cm. What is the area of the trapezoid?A tra

pezoid has two bases that measure 11 cm and 8 cm. The height of the figure is 5 cm. What is the area of the trapezoid?
Mathematics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

47.5

Step-by-step explanation:

A= h·(\frac{b1+b2}{2})

So, 11cm+8cm=19cm

19cm÷2=9.5cm

9.5cm×5cm=47.5cm

Therefore your answer is 47.5cm

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Solve for each equation a)3(x-5)=6 b) 2(x-2/3)=0 c. 4x-5=2-x
Dovator [93]

Answer:

A= 7   B= 0.6  C= 1.4

Step-by-step explanation:

go to m a t h w a y . c o m

8 0
4 years ago
Read 2 more answers
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
aliya0001 [1]

Answer:

A=1500-1450e^{-\dfrac{t}{250}}

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

R_{in} =(concentration of salt in inflow)(input rate of brine)

=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}

R_{out}=(concentration of salt in outflow)(output rate of brine)

=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}

Now, the rate of change of the amount of salt in the tank

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{250}

We solve the resulting differential equation by separation of variables.  

\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}

Taking the integral of both sides

\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}

Recall that when t=0, A(t)=50 (our initial condition)

50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}

4 0
4 years ago
A globe has a diameter of 12 inches. It fits inside a cube-shaped box that has a side length of 12 inches.
AlekseyPX

The volume, rounded to the nearest hundredth of a cubic inch, of the space inside the box that is not taken up by the globe is 823.68 inches³

<h3>Volume of a sphere</h3>

The globe is a sphere. Therefore, the volume is as follows:

volume = 4 / 3 πr³

where

r = 12  /2 = 6 inches

volume = 4 / 3 × π × 6³

volume = 864 / 3 π

volume = 288 × 3.14  = 904.32inches³

<h3>Volume of a cube</h3>

The box is a cube. Therefore,

volume = L³

where

  • L = length

Therefore,

volume = 12³

volume = 1728 inches³

Therefore, the volume of the space inside the box that is not taken up by the globe is as follows:

volume not taken up = 1728 - 904.32 = 823.68 inches³

learn more on volume here: brainly.com/question/15385003

4 0
2 years ago
There are 10 fishes 2 drown 1 runs away and 5 died how many fishes are left?
dezoksy [38]
There are still 10 fish but only 5 are alive because fish don't drown
3 0
3 years ago
Read 2 more answers
I really need help with this, and also if it's Substitution or linear combination and it will be really helpful if you do step b
Delicious77 [7]

Answer:

-2x+y=12

Step-by-step explanation:

linear equation

5 0
3 years ago
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