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2 years ago

Hello everyone i need help with the following question?

Mathematics

1 answer:

Serga [27]2 years ago

6 0

Answer:

9330

Step-by-step explanation:

6x × 8y + 27x - 6y + 3 = 18 × 536 + 81 - 402 + 3 =

9330

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Joshua decides to use synthetic division to find the the quotient of (equation provided above).

tamaranim1 [39]

Answer:

the value to be written is -2

Step-by-step explanation:

The first thing we will do here is to set the divisor equals to zero

The divisor is the denominator

Thus, we have x + 2 = 0

This indicates that x = -2

The term to be written in front of the coefficients is -2

3 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 $m^{3}$ ).

For the density conversion we have to remember that

$1000mL = 1L$

$1L=0.001m^{3}$

$1000g=1Kg$

$\rho=13.5\frac{g}{mL}*\frac{1000mL}{1L}*\frac{1L}{0.001m^{3}}*\frac{1Kg}{1000g} =13500\frac{Kg}{m^{3}}$ .

With this we can finally compute mass:

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

$m=\rho*V$

$m=(13500*10000) Kg$

$m=135000000 Kg$ .

4 0

3 years ago

About how many times would u you expect to get an odd number in 200 trials-I need the answer urgent

ale4655 [162]

Answer:

Step-by-step explanation:

4 0

2 years ago

A delivery truck uses $375 in gasoline during a 3 week period. if the truck is driven 5 days per week, how much is spent on gas

Firdavs [7]

We know that $375 is spent for three weeks period
One week period is 5 days, so for three weeks periods, there are 15 days in total when the truck use the gasoline.

The amount spent on gasoline for one day is $375÷15 = $25

8 0

3 years ago

Last Sunday 1,575 people visited the amusement park 56% of the visitors were adults, 16% were teenagers, and 28% were children a

natulia [17]

<u>Answer:</u>

The number of adults, teenagers out of 1,575 people visited the amusement park are 882 adults and 252 teenagers

<u>Explanation:</u>

Total no of people who visited the amusement park= 1575

Percentage of adults=56%

Percentage of teenagers=16%

Percentage of 12 and under= 28%

Number of adults present= $\frac{56}{100}\times 1575$

=882

Number of teenagers present= $\frac{16}{100}\times 1575$

=252

Hence, 882 adults and 252 teenagers visited the amusement park

5 0

3 years ago