All categories

3 years ago

Solve the following

Mathematics

2 answers:

andre [41]3 years ago

7 0

I got 5 for the first one and 4 for the second one

DENIUS [597]3 years ago

7 0

X =5 X=4 there are the answers hope it helped

You might be interested in

If the mean weight of 4 backfield members on the football team is 222 lb and the mean weight of the 7 other players is 198 lb, w

hoa [83]

To find the Mean or Average you have to add up all numbers it is asking for which in case are 222 and 198.

so you will add 222+198

That will equal 420

Then you will divide with the amount added together which is 2

420/2=210

so the mean of the 11 people is 210 or 210lbs

5 0

3 years ago

A rectangle has a length that is three times the width . The perimeter of the rectangle is 40 feet.what is the area of the recta

charle [14.2K]

Answer:

The area is 15 feet

Step-by-step explanation:

So you are going to need two different equations-

l = w(3)

2l + 2w = 40

Insert the first equation into the second-

2(3w) + 2w = 40

Simplify this equation-

6w + 2w = 40

8w = 40

Divide both sides by 5

w = 5

Solve first equation

l = 5(3)

l = 15

Then find the area-

5 x 15 = 75

7 0

3 years ago

Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =

irga5000 [103]

Answer: The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

8 0

3 years ago

Read 2 more answers

Find the slope of the line passing through the points<br> (-6,-5) and (4,4)

marshall27 [118]

Answer:

$\frac{9}{10}$

Step-by-step explanation:

The slope formula is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Here, $x_{1}$ is -6, $x_{2}$ is 4, $y_{1}$ is -5, and $y_{2}$ is 4.

$m=\frac{4+5}{4+6}$

$m=\frac{9}{10}$

So, the slope of the line passing through the points (-6,-5) and (4,4) is $\frac{9}{10}$ .

3 0

3 years ago

an insurance company sells a policy that pays $50,000.00 in case of accidental death. According to company figures, the rate of

aliya0001 [1]

Answer: $23..50 see attachment for the explanation.

5 0

3 years ago