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2 years ago

Jackson has a collection of 260 coins. How many coins represent 5% of his collection?

Mathematics

2 answers:

krok68 [10]2 years ago

7 0

Answer:

13 coins

Step-by-step explanation:

100% --> 260

1% --> 2.6

5% --> 13

11Alexandr11 [23.1K]2 years ago

4 0

Answer:

13 coins

Step-by-step explanation:

5% of his collection means that we are asking for 5 coins per 100 coins in his current collection. This can be simplified into 5/100 of the collection, which is 1/20. If there are 260 coins in his collection, 1/20 x 260 = 13 coins.

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Find a formula for the described function. an open rectangular box with volume 9 m3 has a square base. express the surface area

Maru [420]

Let

x--------> the length side of the square base

h--------> the height of the box

we know that

the volume of the box is equal to

$V=x^{2} *h\\ V=9\ m^{3}$

$x^{2} *h=9\\\\h=\frac{9}{x^{2} }$

the surface area of the box is equal to

$SA=area\ of\ the\ base+perimeter\ of\ base*height$ (remember that the box is open)

area of the base= $(x^{2})\ m^{2}$

Perimeter of the base= $(4*x)\ m$

height=(h) m

$h=\frac{9}{x^{2}}$

substitute

$SA=x^{2} +4*x*h\\ \\ \\ SA=x^{2} +4*x*\frac{9}{x^{2} } \\ \\ SA=x^{2} +\frac{36}{x}$

we know that

the value of x can not be negative and the denominator can not be zero

therefore

the answer is

the domain of SA is x> 0

the domain is the interval-------------> (0,∞)

6 0

3 years ago

On a number line, the coordinate of X is 52, the coordinate of Y

ASHA 777 [7]

Answer:

Step-by-step explanation:

since the magnitude of xy ≠ zw,

|x-y| = xy = |52-121|= 69, |z-w| = zw = |-21-25| = 46

69 ≠ 46

8 0

3 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$