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3 years ago

40% of students have brown eyes 10 have brown eyes how many atudents in the class

Mathematics

2 answers:

maxonik [38]3 years ago

6 0

50 I believe I think that 40% is 40 + 10 = 50?

castortr0y [4]3 years ago

3 0

Answer:

Step-by-step explanation:

40%x = 10

40/100 x=10

x=10* 100/40

x = 25

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Need help! just need the answer!

Bas_tet [7]

Answer:

Step-by-step explanation:

k=1 ,i THINK

7 0

2 years ago

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Solve the equation for x: c a−x/x−a =5x Answer: , a ≠ − 1/5

meriva

Answer:

$\large\boxed{x=-\dfrac{1}{5}}$

Step-by-step explanation:

$\dfrac{a-x}{x-a}=5x\\\\\dfrac{(a-x)}{-(a-x)}=5x\qquad\text{cancel}\ (a-x)\\\\-1=5x\qquad\text{divide both sides by 5}\\\\-\dfrac{1}{5}=\dfrac{5x}{5}\to\boxed{x=-\dfrac{1}{5}}$

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3 years ago

What would be the answer of 5(x-2)=30 in dustributive property

lisabon 2012 [21]

Answer:

x = 8

Step-by-step explanation:

5(x-2)=30

Distribute

5x- 10 = 30

Add 10 to each side

5x-10+10 = 30+10

5x= 40

Divide by 5

5x/5=40/5

x = 8

3 0

3 years ago

Read 2 more answers

In a recent swim meet, Jason swam 100 meters in 20 seconds.<br> Jason swam ? meters per second

mafiozo [28]

Answer:

5 m/s

Step-by-step explanation:

100 meters divided by 20 seconds

5 0

3 years ago

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The system of equations may have a unique solution, an infinite number of solutions, or no solution. Use matrices to find the ge

Leno4ka [110]

Answer:

Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

$\mid A\mid=5(-1-2)-4(-1-4)+5(1-2)=-15+20-5=0$

Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

4 0

3 years ago