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2 years ago

Rewrite the expression with a rational exponent as a radical expression. (answers in photo)

Mathematics

1 answer:

OleMash [197]2 years ago

3 0

Answer:

The answer is √5

Step-by-step explanation:

That's it

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98 people who were vaccinated were seniors. • 80 people reported a chronic disease or long-term health problem, such as heart or

Fittoniya [83]

Answer:

16.3 is the percent of seniors reported a chronic medical condition chronic is long term recovery acute is short term recovery

Step-by-step explanation:

7 0

3 years ago

The coordinates of a point on a coordinate grid are (−2, 6). The point is reflected across the x-axis to obtain a new point. The

cluponka [151]

The reflection across the x-axis only change the signal of the y coordinate. Hence, the new point is $P'(-2,-6)$ .

4 0

3 years ago

Work out the area of this circle. Take pi to be 3.142 and give your answer to 1 decimal place

Rashid [163]

Step-by-step explanation:

<em>I think you have forgotten to attach the question. </em>

<em>Anyways....</em>

In order to find the area of a circle use the formula:

π $r^{2}$ Where r is the radius

For example the radius is 3cm. The area of the circle will be:

π × $3^{2}$

Since you have taken pi as 3.142...

3.142 × $3^{2}$

3.142 × 9

= 28.278

In order to find the answer to 1 decimal place, check the number next to the number right after the decimal. Is it a number greater than or equal to 5? then add +1 to the number.

= 28.3 $cm^{2}$

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

A company's board of directors wants to form a committee of

NeTakaya

1∪ 2/3.....................

8 0

3 years ago