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3 years ago

5

What is the area of this face

2 answers:

tresset_1 [31]3 years ago

8 0

28in. squared is your answer.

torisob [31]3 years ago

6 0

I think it's 40 not sure though

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Simplify the following expression. −3√84x3

prisoha [69]

Answer:

-18root7

Step-by-step explanation:

-3root 84*3

-3root4*7*3*3

-3(2*3)root7

-18root7

8 0

3 years ago

The sum of two numbers is 74. One number is 6 times less than 4 times the second number. Find the two numbers.

OLEGan [10]

I'll get you started. Let x and y be the two numbers. we have: x + y = 74 (sum is 74) x = 4y - 6 (the number x is six less than 4 times y) Now that you have the equations use an algebraic techniques to eliminate one variable thereby solving for the other. If you need more help add a comment to your question.

3 0

3 years ago

The graph shows the height of a plant over a 2-month growth period. Calculate the rate of change per day.

Mkey [24]

Answer:

14/61 cm per day

Step-by-step explanation:

Use the slope formula $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Plug it in $\frac{16-2}{61-0}$

Simplify $\frac{14}{61}$

3 0

3 years ago

It takes a printer 6 hours to print the class schedules for all of the students in a college. A faster printer can do the job in

poizon [28]

The number of hours that it'll take both printers to do the job is 2.4 hours.

<em>Number of hours used by printer A = 6 hours</em>

<em>Number of hours used by printer B = 4 hours </em>

<em />

Therefore, the number of hours that'll be used by both printers will be calculated thus:

1/6x + 1/4x = 1

Using 12 as the LCM

2x + 3x = 12

5x = 12

Divide both side by 5

5x/5 = 12/5

x = 2.4 hours

Read related link on:

brainly.com/question/25115037

3 0

2 years ago

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview

Anestetic [448]

Let $y_2(t)=tv(t)$ . Then

${y_2}'=tv'+v$

${y_2}''=tv''+2v'$

and substituting these into the ODE gives

$t^2(tv''+2v')+3t(tv'+v)-3tv=0$

$t^3v''+5t^2v'=0$

$tv''+5v'=0$

Let $u(t)=v'(t)$ , so that $u'(t)=v''(t)$ . Then the ODE is linear in $u$ , with

$tu'+5u=0$

Multiply both sides by $t^4$ , so that the left side can be condensed as the derivative of a product:

$t^5u'+5t^4u=(t^5u)'=0$

Integrating both sides and solving for $u(t)$ gives

$t^5u=C\implies u=Ct^{-5}$

Integrate again to solve for $v(t)$ :

$v=C_1t^{-6}+C_2$

and finally, solve for $y_2(t)$ by multiplying both sides by $t$ :

$tv=y_2=C_1t^{-5}+C_2t$

$y_1(t)=t$ already accounts for the $t$ term in this solution, so the other independent solution is $y_2(t)=t^{-5}$ .

6 0

3 years ago