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3 years ago

Can someone help me with this.

Mathematics

1 answer:

Anika [276]3 years ago

4 0

Answer: what method did u use

Step-by-step explanation:

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At right angles.) 4 m 4 m 2 m 2 m 7 m 2 3 m 7 m 11 m

Arisa [49]

Answer:

Step-by-step explanation:

Area of fig 1 = 1 x b = 7 x 4 = 28

Area of fig 2 = l x b = (7-2) x 3 = 5 x 3 = 15

Area of fig 1 = 1 x b = 7 x 4 = 28 (same as ig 1)

Area of whole fig = fig 1 + fig 2 + fig 3 = 28+15+28 = 71

I hope im right !!

6 0

3 years ago

Read 2 more answers

A car has a gasoline

Delvig [45]

Answer:

Yes, look at the explanation:

Step-by-step explanation:

Calculate how many miles a car can drive with 1 gallon: 285 miles ÷ 10 = 28.5 miles with one gallon
Divide 500 by 28.5 to calculate how many gallons are needed: 500 ÷ 28.5 ≈ 17.543 - Therefore yes, it is possible because the car uses about 17.543 gallons of gasoline for 500 miles.

8 0

3 years ago

Pachacha [2.7K]

Answer:

New area: 12ft^2

Step-by-step explanation:

1/3 of 9 is 3

1/3 of 12 is 4

Area = l x w

so 3 x 4 = 12

4 0

3 years ago

N-5÷2=5 help ASAP PLEASE!!!!!

Alinara [238K]

Answer:

7.5 or 15/2 (Whatever form you need to use)

Step-by-step explanation:

N-5 divided by 2 =5

Combine like terms:

-5 divided by 2= -2.5.

N-2.5=5

Add -2.5 from itself, then do the same to the other side:

N-2.5=5

+2.5=+2.5 (Adding 2.5 to -2.5 cancels it out)

N=7.5

6 0

3 years ago

Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

7 0

4 years ago