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4 years ago

What is 36/4 as a mixed number

Mathematics

1 answer:

Yuliya22 [10]4 years ago

3 0

It cannot be expressed as a mixed fraction. Dividing 36 by 4 gives 9 which is a whole number.

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Solve: log4(7t + 2) = 2

IRISSAK [1]

Answer:

The required value of t is equal to 2

Step-by-step explanation:

We need to solve the following expression :

$\log_4(7t+2)=2$

Now, using the base rule :

$2=\log_4(4^2)\\\\\implies 2= \log_4(16)$

Now, using this value of 2

$\log_4(7t+2)=2\\\\\implies\log_4(7t+2)=\log_4(16)\\\\\implies 7t+2=16\\\\\implies 7t=16-2\\\\\implies 7t=14\\\\\implies t = 2$

Hence, The required value of t is equal to 2

3 0

3 years ago

Area codes have three digits. The first digit must not be 0 or 1, the middle digit must be 0 or 1, and the last digit must not b

disa [49]

There are 10 digits 0 1 2 3 4 5 6 7 8 9 if we can't use 0 and 1 then there are only 8 choices for the first digit and only 2 choices for the second digit and 10 for the third so there are 8*2*10 =160 possibilities. I think ???

4 0

3 years ago

(-3x - 7) + (4x + 7) =

Alex73 [517]

Answer:

Not sure if you want it factored or simplified

(-3x - 7) + (4x + 7) = x

Cannot factor

Step-by-step explanation:

8 0

3 years ago

Calculate the number of kilograms of mercury required to fill a pool that is 20.0 m x 50.0 m x 10.0 m. The density of mercury is

malfutka [58]

Answer:

$m=135000000 Kg$

Step-by-step explanation:

To find the number of kilograms of mercury we need to find how to relate density, mass and, volume. For this we shall recall the density formula:

$\rho=\frac{m}{V}$

where $\rho$ is the density, $m$ is the mass and, $V$ is the volume.

We have the density and want to compute the mass so now we want to know the volume of the pool.

The volume of a rectangular pool is given by the fomula:

$V= (length)(heigh)(width)$ .

So for our pool

$V= (20)(50)(10)$ .

$V= 10000m^{3}$ .

Our density is in $g/mL$ , so the last thing we need to do before computing the mass is to express the density in $Kg/m^{3}$ (this is because we want our mass in $Kg$ and, we have our volume in $m^{3}$ ).