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brilliants [131]
3 years ago
8

Find the area of the trapezoid? and no links ​

Mathematics
2 answers:
vovikov84 [41]3 years ago
7 0
Add all the sides n u will get ur answer
MAVERICK [17]3 years ago
5 0

Area = 1/2 (base 1 + base 2) height

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Help me pls this one is hard
ella [17]

Answer:

2-120

4-240

7-420

Step-by-step explanation:

<em>there are 60 people in each tour group so you multiply 60 by the number of tour groups.</em>

ordered pairs:

<em>(2,120)</em>

<em>(4,240)</em>

<em>(7,420)</em>

hope this helped you!

have a great day!!

7 0
3 years ago
Read 2 more answers
This math promblem!!!!
fiasKO [112]
1. You convert all the numbers into decimals.
   a. For 8 1/9 you multiply 8x9 and add the numerator which in this case is one, so the equation would be 8x9=72     then 72+1= 73
   b. For 81/10 I used a calculator for accuracy and I just divided 81 by 10 because the fraction line can also be used as a division sign. For this I got 8.1
2. Now I looked at all the numbers I had including the fractions I converted to decimals... 8.115, 8.55, 73, and 8.1
3. Lastly, I put the numbers in order from least to greatest: 8.1, 8.115, 8.55, and 73
4. In order to figure out which one is the smallest and largest, I just added zeros on the end of the numbers so they would all be the same: 8.1-->8.100, 8.115 I kept the same because it already had 3 decimal places, 8.55--> 8.550, and 73--> 73.000
5. Then i could tell which number was the largest by the decimal place numbers.
**Hope this was helpful... It's kind of hard to explain online but hopefully you have a better understanding of how to do it!**
6 0
3 years ago
Read 2 more answers
How many cases does he hear per hour?
IRISSAK [1]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Help me answer these questions please 15, 19 and 20 please​
SpyIntel [72]

Answer:

15) K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]

19) P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]

20) Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)

Step-by-step explanation:

We are to find the derivative of the questions pointed out.

15) K(t) = 5(5^(t)) - 2(3^(t))

Using implicit differentiation, we have;

K'(t) = 5[5^(t)•In 5] - 2[3^(t)•In 3]

19) P(w) = 2e^(w) - (2^(w))/5

P'(w) = 2e^(w) - (1/5)[2^(w)•In 2]

20) Q(W) = 3w^(-2) + w^(-2/5) - w^(¼)

Q'(w) = -6w^(-2 - 1) + (-2/5)w^(-2/5 - 1) - ¼w^(¼ - 1)

Q'(w) = -6w^(-3) - (2/5)w^(-7/5) - ¼w^(-¾)

7 0
3 years ago
Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution
KIM [24]

Answer:

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be y=e^{mx}

Then, y'=me^{mx}   and   y''=m^2e^{mx}

\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0

\Rightarrow m^2-m-20=0

\Rightarrow m^2-5m+4m-20=0

\Rightarrow m(m-5)+4(m-5)=0

\Rightarrow (m-5)(m+4)=0

\Rightarrow m=-4,5

Therefore the complementary function is = c_1e^{-4x}+c_2e^{5x}

Therefore e^{-4x} and e^{5x} are fundamental solution of the given differential equation.

If y_1 and y_2 are the fundamental solution of differential equation, then

W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0

Then  y_1 and y_2 are linearly independent.

W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|

                    =e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})

                    =5e^x+4e^x

                   =9e^x\neq 0

Therefore  e^{-4x} and e^{5x} are linearly independent, since W(e^{-4x},e^{5x})=9e^x\neq 0

Let the the particular solution of the differential equation is

y_p=v_1e^{-4x}+v_2e^{5x}

\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx

and

\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx

Here y_1= e^{-4x}, y_2=e^{5x},W(e^{-4x},e^{5x})=9e^x ,and  R(x)=0

\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx

       =0

and

\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx

       =0

The the P.I = 0

The general solution of the differential equation is

y=c_1e^{-4x}+c_2e^{5x}

7 0
3 years ago
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