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3 years ago

9

Simone takes dance lessons. Her shoes cost $150 per year, her tights and leotards cost $70 per year, her hair accessories cost $

15 per year, and the dance lessons cost $1,000 per year. How much will Simone pay for five years of dance lessons?

1 answer:

KATRIN_1 [288]3 years ago

6 0

Ok so all you do is add 150+70+15+1000. That equals to $1235

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£100 in the ratio 3:2

trasher [3.6K]

Answer:

66.66 (other currency)

Step-by-step explanation:

If every £3 represents 2 of another currency, you just need to figure out how many groups of £3 are in £100 (so £100/£3 = 33.33). Now multiply this value by 2 to get the equivalent of £100 in another currency.

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3 years ago

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31.205*4.97/1.925 correct to 1 significant number

Sonja [21]

Answer:80

Step-by-step explanation:

31.205= 31

4.97=5

1.925=2

31*5/2=155/2=77.5=80 (1 significant figure)

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2 years ago

Evaluate the determinant of the matrix.<br> -4 5 6<br> 0 4 4<br> -2 -5 4<br> I need the work plz

FromTheMoon [43]

Answer:

-136

Step-by-step explanation:

We have to find the determinant of the following matrix:

$\left[\begin{array}{ccc}-4&5&6\\0&4&4\\-2&-5&4\end{array}\right]$

We can find the determinant by expanding via 1st column. i.e. by taking each element of 1st column and multiplying it by its co-factor matrix as shown below:

det $\left[\begin{array}{ccc}-4&5&6\\0&4&4\\-2&-5&4\end{array}\right]$

= $(-4 \times det \left[\begin{array}{cc}4&4\\-5&4\end{array}\right]) - (0 \times (-4 \times det \left[\begin{array}{cc}5&6\\-5&4\end{array}\right]))+ ((-2) \times det\left[\begin{array}{cc}5&6\\4&4\end{array}\right])\\\\ =-4 \times (16 + 20)-(0)+(-2 \times 20-24)\\\\ =-4(36)+(-2(-4))\\\\ =-144+8\\\\ =-136$

The notation det() stands for determinant of the matrix.

Therefore, the determinant of the given matrix is -136

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3 years ago

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Kitty [74]

Answer:

The values of $a$ and $b$ so that the two linear equations have infinite solutions are $3$ and $\frac{3}{2}$ , respectively.

Step-by-step explanation:

Two linear equations with two variables have infinite solution if and only if they are<em> linearly dependent</em>. That is, one linear equation is a multiple of the other one. Let be the following system of linear equations:

$2\cdot x + y = 6$ (1)

$a\cdot x + b\cdot y = 9$ (2)

The following condition must be observed:

$r = \frac{a}{2} = \frac{b}{1} = \frac{9}{6}$ (3)

After some quick operations, we find the following information:

$r = \frac{3}{2}$ , $a = 3$ , $b = \frac{3}{2}$

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Phase shift = 2
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