All categories

4 years ago

. Name the place value of the digit 3 in the number 3,240,987

Mathematics

1 answer:

Sonbull [250]4 years ago

3 0

The place value of the digit 3 is millions.

Hope this helps!

You might be interested in

I need help it's due 1:25 (eastern time

Nimfa-mama [501]

Answer:

- 7/8

Step-by-step explanation:

m = Δy/Δx = (10-3)/(2-10) = - 7/8

5 0

3 years ago

Read 2 more answers

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

Which expression can be used to find the surface area of the following square pyramid?

kodGreya [7K]

Answer:

Step-by-step explanation:

expression for surface area = a² + a √(a²/4+h²)

=4² + 2(4) √(4²/4+5²)

=16 + 8 √(16/29)

= 24 √0.55

= 20.65

6 0

3 years ago

PLEASE HELP GIVING BRAINLY!!! (if you could at least help me with 1 or even both I’ll give brainly!!)

Fiesta28 [93]

Answer:

1. 4

Step-by-step explanation:

24 - 20, if you add the the square then the rectangle, then subtract

7 0

3 years ago

NEED HELP ASAP!!!!!!!!! (WILL GIVE BRAINLIEST)

Ipatiy [6.2K]

Answer:

i think B

hope this helps

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers