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2 years ago

Round the number to the indicated place. $13.59562 to cents

Mathematics

1 answer:

Aleks04 [339]2 years ago

7 0

Answer:

look below

Step-by-step explanation:

Lets round the 0.59562 first

0.59562 rounded to the nearest hundreth (because thats how cents go) rounds up to .60

there is 100 cents in a dollar

there is 13 dollars

13 * 100 is 1300

1300 + 60

1360 cents

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solve each equation by finding square roots. If the equation has no real number solution, write no solution. 64b^2=16

ikadub [295]

B=1/4 hope it helps sorry if I’m wrong

6 0

3 years ago

Marilyn collects old dolls. She purchases a doll for $450. Research shows the doll's value with increase by 2.5% each year.

Irina-Kira [14]

Answer:

$\$737.38$

Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

$y=a(1+r)^x$

where

y is the value of the doll

x is the number of years

a is the initial value

r is the rate of change

we have

$a=\$450\\r=2.5\%=2.5/100=0.025$

substitute

$y=450(1+0.025)^x$

$y=450(1.025)^x$

What will be the value of the doll in 20 years?

For x=20 years

substitute in the equation

$y=450(1.025)^{20}=\$737.38$

8 0

4 years ago

A system of equations is given by 4x1 − 2x2 − 7x3 = 4 2x1 + 12x2 + 3x3 = −5 −x1 + 6x2 + 2x3 = 15 i) Write down the augmented coe

Bingel [31]

Answer:

i) The augmented matrix for this system is

$\left[\begin{array}{ccc|c}4&-2&-7&4\\2&12&3&-5\\-1&6&2&15\end{array}\right]$

ii) The reduced row echelon form

$\text{rref}(A)=\left[ \begin{array}{cccc} 1 & 0 & 0 & - \frac{692}{65} \\\\ 0 & 1 & 0 & \frac{423}{130} \\\\ 0 & 0 & 1 & - \frac{493}{65} \end{array} \right]$

iii) The set of solutions are

$x_1$ = -692/65

$x_2$ = 423/130

$x_3$ = -493/65

Step-by-step explanation:

i) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

We have the following system of equations:

$4x_1 -2x_2 - 7x_3 = 4 \\2x_1 + 12x_2 + 3x_3 = -5\\ -x_1 + 6x_2 + 2x_3 = 15$

Here is the augmented matrix for this system.

$\left[\begin{array}{ccc|c}4&-2&-7&4\\2&12&3&-5\\-1&6&2&15\end{array}\right]$

ii) To find the reduced row echelon form of

$A=\left[ \begin{array}{cccc} 4 & -2 & -7 & 4 \\\\ 2 & 12 & 3 & -5 \\\\ -1 & 6 & 2 & 15 \end{array} \right]$

you must:

Divide row 1 by 4 $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{cccc} 1 & - \frac{1}{2} & - \frac{7}{4} & 1 \\\\ 2 & 12 & 3 & -5 \\\\ -1 & 6 & 2 & 15 \end{array} \right]$

Subtract row 1 multiplied by 2 from row 2 $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{cccc} 1 & - \frac{1}{2} & - \frac{7}{4} & 1 \\\\ 0 & 13 & \frac{13}{2} & -7 \\\\ -1 & 6 & 2 & 15 \end{array} \right]$

Add row 1 to row 3 $\left(R_3=R_3+R_1\right)$

$\left[ \begin{array}{cccc} 1 & - \frac{1}{2} & - \frac{7}{4} & 1 \\\\ 0 & 13 & \frac{13}{2} & -7 \\\\ 0 & \frac{11}{2} & \frac{1}{4} & 16 \end{array} \right]$

Divide row 2 by 13 $\left(R_2=\frac{R_2}{13}\right)$

$\left[ \begin{array}{cccc} 1 & - \frac{1}{2} & - \frac{7}{4} & 1 \\\\ 0 & 1 & \frac{1}{2} & - \frac{7}{13} \\\\ 0 & \frac{11}{2} & \frac{1}{4} & 16 \end{array} \right]$

Add row 2 multiplied by 1/2 to row 1 $\left(R_1=R_1+\left(\frac{1}{2}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & - \frac{3}{2} & \frac{19}{26} \\\\ 0 & 1 & \frac{1}{2} & - \frac{7}{13} \\\\ 0 & \frac{11}{2} & \frac{1}{4} & 16 \end{array} \right]$

Subtract row 2 multiplied by 11/2 from row 3

$\left[ \begin{array}{cccc} 1 & 0 & - \frac{3}{2} & \frac{19}{26} \\\\ 0 & 1 & \frac{1}{2} & - \frac{7}{13} \\\\ 0 & 0 & - \frac{5}{2} & \frac{493}{26} \end{array} \right]$

Multiply row 3 by −2/5 $\left(R_3=\left(- \frac{2}{5}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & - \frac{3}{2} & \frac{19}{26} \\\\ 0 & 1 & \frac{1}{2} & - \frac{7}{13} \\\\ 0 & 0 & 1 & - \frac{493}{65} \end{array} \right]$

Add row 3 multiplied by 3/2 to row 1 $\left(R_1=R_1+\left(\frac{3}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & - \frac{692}{65} \\\\ 0 & 1 & \frac{1}{2} & - \frac{7}{13} \\\\ 0 & 0 & 1 & - \frac{493}{65} \end{array} \right]$

Subtract row 3 multiplied by 1/2 from row 2 $\left(R_2=R_2-\left(\frac{1}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & - \frac{692}{65} \\\\ 0 & 1 & 0 & \frac{423}{130} \\\\ 0 & 0 & 1 & - \frac{493}{65} \end{array} \right]$

The answer is

$\text{rref}(A)=\left[ \begin{array}{cccc} 1 & 0 & 0 & - \frac{692}{65} \\\\ 0 & 1 & 0 & \frac{423}{130} \\\\ 0 & 0 & 1 & - \frac{493}{65} \end{array} \right]$

iii) The set of solutions are

$x_1$ = -692/65

$x_2$ = 423/130

$x_3$ = -493/65

5 0

3 years ago

Andres buys 3 boxes of markers. Each box has the same number of markers. Andres now has 15 markers. Write and solve an equation

irinina [24]

Answer:

The answer is five

Step-by-step explanation:

Because 5 times 3 equals 15.

That's just basic multiplying. (*^*)

Have a good day!

3 0

3 years ago

Read 2 more answers

Find the nth term of this quadratic sequence 7,16,25,34,43

GalinKa [24]

Answer:

Step-by-step explanation:

everytime it adds by 9:

7+9=16

16+9=25

25+9=34

34+9=43

<u>43+9=52</u>

6 0

3 years ago