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2 years ago

Solve the following equation for p: 6/p=x+a

1 answer:

6 0

$\dfrac 6p = x+a\\\\\implies \dfrac 6p \times\dfrac 16 = (x+a) \times\dfrac 16 ~~~~~~;\left[\text{Multiply both sides by}~ \dfrac 16 \right]\\\\\implies \dfrac 1p = \dfrac{x+a}6\\\\\implies p= \dfrac{6}{x+a}~~~~~~~~~~~~~~~~~~~~;[\text{Cross multiply}]$

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It takes 34 pounds of seed to completely plants 5 acre field. How many pounds of seed are needed per acre?

Nataly [62]

Answer:

Step-by-step explanation:

Since 34 Lbs of seed is needed for 5 Acre's, You will need two find 1/5th of 34.

1/5 in decimal form is 0.2

34*0.2= 6.8

In order to cover 1 Acre of land you will need 6.8 lbs of seed.

Hope this helps!

6 0

3 years ago

Let x represent the number of television show episodes that are taped in a season. Enter an expression for the number of episode

Vinil7 [7]

Answer:

x=8 ???

Step-by-step explanation:

??? not 100% sure but im pretty sure ;w;

8 0

3 years ago

Read 2 more answers

What is the value of x? (SOMEONE PLEASE HELP ME OUT BRO PLEASEEEEE IGNORE THE 9 smh)

max2010maxim [7]

Answer:

x=28

Step-by-step explanation:

Sum of all the angles in the triangle add up to equal 180

90 + x + 6 + 2x = 180

combine like terns

96 + 3x = 180

subtract 96 from each side

180-96=84

96-96 cancels out

3x=84

divide each side by 3

84/3=28

3/3 cancels out

x=28

The1AndOnlyMarkus

6 0

3 years ago

Read 2 more answers

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. N(10) corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

3 years ago

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5, find the ratio of the smaller number to the larger nu

IgorC [24]

Answer:

$\frac{1}{9}$

Step-by-step explanation:

Let the numbers be x,y, where x>y

The geometric mean is

$\sqrt{xy}$

The Arithmetic mean is

$\frac{x + y}{2}$

The ratio of the geometric mean and arithmetic mean of two numbers is 3:5.

$\frac{ \sqrt{xy} }{ \frac{x + y}{2} } = \frac{3}{5}$

We can write the equation;

$\sqrt{xy} = 3$

$xy = 9 - - - (2)$

and

$\frac{x + y}{2} = 5$

$x + y = 10 - - - (2)$

Make y the subject in equation 2

$y = 10 - x - - - (3)$

Put equation 3 in 1

$x(10 - x) = 9$

$10x - {x}^{2} = 9$

${x}^{2} - 10x + 9 = 0$

$(x - 9)(x - 1) = 0$

$x =1 \: or \: 9$

When x=1, y=10-1=9

When x=9, y=10-9=1

Therefore x=9, and y=1

The ratio of the smaller number to the larger number is

$\frac{1}{9}$

3 0

3 years ago