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2 years ago

10.4-7.4 than write it as an improper fraction

Mathematics

2 answers:

Stella [2.4K]2 years ago

6 0

1. 4/3
2.67/10
3.19/8
4.23/5
5.15/4
6. 3/4
7 i can not see 7.

I hope this helps!!!!!!!!!!!

egoroff_w [7]2 years ago

3 0

6) 3/4. 5)30/8. 3)19/8. 1)16/12

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This is due at 8:00PM please someone help

krek1111 [17]

10. A 11. D 12. A (wasn’t too sure about question 12)

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2 years ago

Given the formula for an arithmetic sequence f(7) = f(6) + 4 written using a recursive formula, write the sequence using an arit

natita [175]

Answer:

Of(7) = f(1) + 24

Step-by-step explanation:

Since this Arithmetic Sequence can be written recursively as a function, then we can write the whole sequence, by adding the common difference to the previous function. So writing it as an Arithmetic formula is (placing an example, with a common difference of 4 units):

$\left.\begin{matrix}n &1&2&3&4&5&6 &7\\f(n)&2&6&10 & 14&18&22&26 \end{matrix}\right|\\f(n)=f(n-1)+d\:\: (Recursive \: Formula)\Rightarrow a_{n}=a_{n-1}+4\: \: \: (Arithmetic\: Formula)\\Of(7)=f(1)+6*4\Rightarrow a_{7}=a_{1}+(7-1)4\Rightarrow a_{7}=a_{1}+24\\$

6 0

3 years ago

Read 2 more answers

What is the area of the triangle below (in square units)?

Helga [31]

Answer:

It is 50 cm squared hxb/2=a

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2 years ago

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 $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

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,

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