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2 years ago

I need help again don't get it

Mathematics

2 answers:

Dmitry [639]2 years ago

3 0

Answer:

x=45

Step-by-step explanation:

70+65=135

180-135=45

Alenkasestr [34]2 years ago

3 0

Answer:

The sum of all the angle in a triangle is 180

so 70+65 = 135

then you do 180-135=45 = x

and thats you answer

hope that answers your question

Step-by-step explanation:

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Answer the following questions Question one Are the following rational, irrational or neither? Please HELP ASAP

Alex777 [14]

Answer:

1.1) Rational

1.2) Rational

1.3) Irrational

1.4) Rational

1.5) Rational

1.6) Irrational

Step-by-step explanation:

Rational means it can be represented as a fraction where the numerator and denominator are whole numbers and vice versa for irrationals.

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2 years ago

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Caleb was out at a restaurant for dinner when the bill came. His dinner came to $13. He wanted to leave a 11% tip. How much was

frozen [14]

Answer: 13+ 1.43 witch is 14.43

Step-by-step explanation: 11% of 13 is .11 times 13 witch is 1.43 so add them and you get the answer of 14.43

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2 years ago

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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 $R(x)=0$

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

and

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

The the P.I = 0

The general solution of the differential equation is

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

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2 years ago

7) CJ was painting his apartment. In 15 minutes,

igomit [66]

y = 12x represents the gallons of paint used in x hours.

Step-by-step explanation:

CJ uses 3 gallons of paint in 15 minutes,

15 minutes = 3 gallons

1 minutes = $\frac{3}{15} = \frac{1}{5} \ gallons\\$

1 hour = 60 minutes

Gallons of paint used in 1 hour;

1 hour = $\frac{1}{5}*60=12\ gallons$

CJ uses 12 gallons of paint in 1 hour.

For x hours = 12*x = 12x

Gallons of paints used = y

y = 12x represents the gallons of paint used in x hours.

Keywords: linear equation, multiplication

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The answer would be 3

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