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3 years ago

Please Fraction help & explain

Mathematics

1 answer:

Anika [276]3 years ago

8 0

Answer:

5/2

Step-by-step explanation:

Convert both fractions into improper fractions:

- To convert these fractions into an improper fraction you need to multiple the denominator with the mixed numeral/whole number. Then add that amount to the numerator. Do this for both fractions. This is what it should look like.

38/6 - 23/6.

The you subtract it

- 38/6 - 23/6 = 15/6

15/6 is your answer

If you want to simplify it the fraction it will be :

- 5/2

You might be interested in

Dilate the the triangle, scale factor = 3 (positive 3, ignore the

HACTEHA [7]

Answer:

The rule of dilation is $P'(x,y) = 3\cdot P(x,y)$ .

The vertices of the dilated triangle are $A'(x,y) = (-21, -18)$ , $B'(x,y) = (-15,-10)$ and $C'(x,y) = (-3,-15)$ , respectively.

Step-by-step explanation:

From Linear Algebra, we define the dilation by the following definition:

$P'(x,y) = O(x,y) + k\cdot[P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation, dimensionless.

$k$ - Scale factor, dimensionless.

$P(x,y)$ - Original point, dimensionless.

$P'(x,y)$ - Dilated point, dimensionless.

If we know that $O(x,y) = (0,0)$ , $k = 3$ , $A(x,y) = (-7,-6)$ , $B(x,y) = (-5,-2)$ and $C(x,y) =(-1,-5)$ , then dilated points of triangle ABC are, respectively:

$A'(x,y) = O(x,y) + k\cdot [A(x,y)-O(x,y)]$ (2)

$A'(x,y) = (0,0) + 3\cdot [(-7,-6)-(0,0)]$

$A'(x,y) = (-21, -18)$

$B'(x,y) = O(x,y) + k\cdot [B(x,y)-O(x,y)]$ (3)

$B'(x,y) = (0,0) + 3\cdot [(-5,-2)-(0,0)]$

$B'(x,y) = (-15,-10)$

$C'(x,y) = O(x,y) + k\cdot [C(x,y)-O(x,y)]$ (4)

$C'(x,y) = (0,0) +3\cdot [(-1,-5)-(0,0)]$

$C'(x,y) = (-3,-15)$

The rule of dilation is:

$P'(x,y) = 3\cdot P(x,y)$ (5)

The vertices of the dilated triangle are $A'(x,y) = (-21, -18)$ , $B'(x,y) = (-15,-10)$ and $C'(x,y) = (-3,-15)$ , respectively.

7 0

3 years ago

For Jane's Uber business, She charges 5$ intial fee and $0.10 a mile. Joe's Uberbusiness charges $4 initial fee and $.0.20 a mil

marusya05 [52]

jane- y=0.10x+5

joe- y=0.20x+4

7 0

4 years ago

Solve the system of equations.

emmainna [20.7K]

$x = \frac{ 419 }{ 113 } ~,~y = -\frac{ 208 }{ 113 } ~,~z = \frac{ 21 }{ 113 }$

<h2>Explanation:</h2>

We have the following system of three linear equations:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\5~ x&+~~8~ y&+~~~~ z&~=~4\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}$

Let's use elimination method in order to get the solution of this system of equation, so let's solve this step by step.

Step 1: Multiply first equation by $-5/2$ and add the result to the second equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\4~ x&+~~5~ y&+~~2~ z&~=~6\end{array}$

Step 2: Multiply first equation by −2 and add the result to the third equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&-~~~3~ y&-~~~62~ z&~=~-6\end{array}$

Step 3: Multiply second equation by −32 and add the result to the third equation. So we get:

$\begin{array}{ cccc }2~ x&+~~4~ y&+~~32~ z&~=~6\\&-~~~2~ y&-~~~79~ z&~=~-11\\&&+~~\frac{ 113 }{ 2 }~ z&~=~\frac{ 21 }{ 2 }\end{array}$

Step 4: solve for z.

$\begin{aligned} \frac{ 113 }{ 2 } ~ z & = \frac{ 21 }{ 2 } \\ z & = \frac{ 21 }{ 113 } \end{aligned}$

Step 5: solve for y.

$\begin{aligned}-2y-79z &= -11\\-2y-79\cdot \frac{ 21 }{ 113 } &= -11\\y &= -\frac{ 208 }{ 113 } \end{aligned}$

Step 6: solve for x by substituting $y=-\frac{208}{113}$ and $z = \frac{21}{113}$ into the first equation:

$2x+4(-\frac{208}{113})+32(\frac{21}{113})=6 \\ \\ 2x-\frac{832}{113}+\frac{672}{113}=6 \\ \\ 2x=6+\frac{832}{113}-\frac{672}{113} \\ \\ 2x=\frac{838}{113} \\ \\ x=\frac{319}{113}$