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3 years ago

Manolo and Annalisa make handbags. Manolo made 16 handbags the first day and 12 each day after

Mathematics

1 answer:

saw5 [17]3 years ago

4 0

Based on the amount being made by Manolo and Annalisa every day, the amount both made on the various days were:

Annalisa made 8 bags on Day 1
Manolo had made 40 bags by Day 3

Annalisa is said to have made 8 bags on the first day so this is the number of bags that should be put in the table for having been made in the first day.

Manolo made 16 bags on the first day and then 12 bags each day after that. The total number of bags made by the third day by Manolo would be:

= 16 + 12 + 12

= 40 bags

In conclusion, Annalisa made 8 bags on Day 1 and Manolo had made 40 bags on the 3rd day.

Find out more about this at brainly.com/question/11864663.

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What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

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Sidana [21]

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Which choice shows the coordinates of C’ if the trapezoid is reflected across the y-axis?

suter [353]

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. How do we write simple expressions using long & synthetic division? Explain

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Hi there,

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