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3 years ago

Find the product of 4025 multiply 5 by using properties

Mathematics

1 answer:

strojnjashka [21]3 years ago

6 0

Answer:

Change 4020 to 4000 + 25.

Then use the distributive property.

4025 * 5 = (4000 + 25) * 5 = 4000 * 5 + 25 * 5 = 20,000 + 125 = 20,125

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Igoryamba

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3 years ago

Jerry purchased three pens for $1.19 each and six notebooks for $2.29 each. How many did Jerry spend altogether?

Masteriza [31]

Hi!

We will find the sum you are looking for like this:

3 * 1,19$ + 6 * 2,29$ = 3,57$ + 13,74$ = 17,31$.

Hope this helps!

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3 years ago

Given that, 1 Chinese yuan = £0.11, convert £460 into Chinese yuan. Give your answer to the nearest yuan.

Ugo [173]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

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,

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