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3 years ago

Question 14 LAST ONE will mark as BRAILEST

Mathematics

2 answers:

Delicious77 [7]3 years ago

5 0

Answer: C

Step-by-step explanation:

rodikova [14]3 years ago

4 0

The answer is indeed c

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Need help to answer

Korolek [52]

Remember, you can do anything to an equation as long as you do it to both sides

times 4 both sides because we hate fractions
3x+8=16x-4
minus 3x both sides
8=13x-4
add 4 both sides
12=13x
divide both sides by 13
12/13=x
x=12/13

4 0

3 years ago

This is giving me a hard time

kirill [66]

Answer:

x= 9,-2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Consider the differential equation

Ainat [17]

Answer:

$W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x$

Step-by-step explanation:

Let $y=e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=\frac{1}{2}e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y''=\frac{1}{4}e^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (\frac{1}{4}e^{\frac{x}{2}} \right )-4\left ( \frac{1}{2}e^{\frac{x}{2}}\right )+e^{\frac{x}{2}}\\=e^{\frac{x}{2}}-2e^{\frac{x}{2}}+e^{\frac{x}{2}}\\=2e^{\frac{x}{2}}-2e^{\frac{x}{2}}\\=0$

So, $y=e^{\frac{x}{2}}$ is the solution of the given equation.

Now, let $y=xe^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=e^{\frac{x}{2}}+\frac{x}{2}e^{\frac{x}{2}}=e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )$

Differentiate with respect to $x$

$y''=\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{2}e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )=e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}} \right )-4\left ( e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0$

To find: $W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )$

Solution:

Let $u=e^{\frac{x}{2}}\,,\,v=xe^{\frac{x}{2}}$

$W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x$

5 0

3 years ago

In a certain town, 42% of voters favor a given ballot measure. For groups of 30 voters, find the

alex41 [277]

Answer:

7.3 voters

Step-by-step explanation:

This scenario follows a binomial distribution with the probability of success (voting in favor of the ballot) p =0.42, and the probability of failure (voting against the ballot), p-1 = 0.58. If the number of events is n=30. The variance is given by:

$V= n*p*(1-p)\\V= 30*0.42*(1-0.42)\\V=7.3\ voters$

The variance for the number who favor the measure is 7.3 voters.

6 0

3 years ago

Are these angles Supplementary or Complementary

gregori [183]

Answer:

they are suplementary

Step-by-step explanation:

because no right angle and less then 90 degrees

7 0

3 years ago

Read 2 more answers