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3 years ago

10

14.51 less than the product of 26 and x

1 answer:

Mekhanik [1.2K]3 years ago

6 0

<em>the</em><em> </em><em>pro</em><em>duct</em><em> </em><em>of</em><em> </em><em>2</em><em>6</em><em> </em><em>and</em><em> </em><em>x</em><em> </em><em>is</em><em> </em><em>wri</em><em>tten</em><em> </em><em>as</em>

<em>2</em><em>6</em><em> </em><em>times</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>2</em><em>6</em><em>x</em>

<em> $= 14.51 < 26x$ </em>

<em> $= 1451 \times {10}^{ - 2} < 26x$ </em>

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3+7+11+15+19+23+27<br>1.t1=<br>s1=<br>express t1 in terms of s1

icang [17]

So it would be 3s + 7s + 11s + 15s + 19s + 23s + 27s. Hope this helps

8 0

2 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago

An electronic device that previously sold for $21.00 has been reduced to $17.43. The price reduction, rounded to the near east w

svet-max [94.6K]

The percentage of reduction is 17%

<u>Explanation:</u>

Original price = $21

Reduced price = $17.43

Percentage of reduction = ?

Difference in price = $21 - $17.43

= $3.57

Percentage of reduction = $\frac{difference}{original} X 100$

$= \frac{3.57}{21} X 100\\\\= 17$

Therefore, percentage of reduction is 17%

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3 years ago

2-2x2-(39 divided by 1)+2+9

artcher [175]

-30 is the answer for your question

6 0

3 years ago

Explain how you know that x^2-8x+20 is always positive. <br> Please help.

IRISSAK [1]

Answer:

if you rearrange to complete the square, you get (x^2-4)^2 +4

and seeing as anything squared will always be positive or zero, the lowest possible value for (x^2-4)^2 is 0, when x = 4

and 0 + 4 = 4, which is greater than 0, so positive

Step-by-step explanation:

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3 years ago