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2 years ago

A newspaper prints that the average square footage of a house in Cleveland is significantly less than 2,379 square feet.

Mathematics

2 answers:

tatyana61 [14]2 years ago

6 0

Answer: i’m confused on what the question is

Step-by-step explanation:

kvv77 [185]2 years ago

4 0

Answer:

thats not true

Step-by-step explanation:

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In the diagram, AE = 2AC and ∠BAC ≅ ∠DAE. What additional information is necessary to prove that ΔABC is similar to ΔADE, using

ss7ja [257]

Since you want to use the SAS theorem, you must find sides that are either side of angles BAC and DAE. You have already made use of sides AE and AC, so the other sides you need to choose are AB and AD. The appropriate relationship for similarity is ...

... AD = 2AB

since you want the sides of triangle ADE to be twice then length of those in triangle ABC.

7 0

3 years ago

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Write an equation of a line in slope intercept form that is perpendicular to y = -4x -2 and passes through the point (-16, -11).

NISA [10]

Y = -4x - 2
P (-16, -11)
a = -4, b = 0, c = -2
m = -a/b
m = -(-4)/(-2)
m = 4/(-2)
m = -2

y-yo = m(x-xo)
y-(-11) = -2[x-(-16)]
y+11 = -2(x+16)
y+11 = -2x-32
2x+y+11+32 = 0
2x+y+43 = 0

A outra questão é semelhante.

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2 years ago

A football coach is trying to decide: When a team is ahead late in the game,

brilliants [131]

Answer:

I play football and my coach will do this: Play a "prevent* defense that guards against long gains but makes short

gains easier.

Step-by-step explanation:

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1 year ago

In order for you to answer the question correctly, please use the following image down below:

adell [148]

Answer:

Step-by-step explanation:

24^2 = (12)(12+x)

576 = 144 + 12x

432 = 12x

36 = x

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2 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

The approximate difference in the half-lives of the isotopes is 66 days.

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2 years ago

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