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Nezavi [6.7K]
3 years ago
5

ANSWER ASAPPP PLEASEEEEEE

Mathematics
1 answer:
Law Incorporation [45]3 years ago
5 0
The following sum is irrational. √16 is rational because it's perfect square, the result is 4. But √8 is not rational because it's not perfect square.

The answer is second option
You might be interested in
Amy bought a diamond ring for $6,000. If the value
Temka [501]

Answer: $13,308

5.8% of 6000 is 348. 348 times 21 is $7,308. add 6,000 to 7,308 and you get $13,308

Step-by-step explanation: amy likes cheap rings LOL

3 0
3 years ago
Read 2 more answers
two stores sell the same computer for the same orginal price. store a advertises that the computer is 25% off the orginal price.
Gnoma [55]
That means that the discounts are the same
therfor 25% of original=180
multiply both sides by 4 or divide both sides by 0.25
original=720




the equation is
0.25 times original=180


and the original price is $720
4 0
3 years ago
Is this the correct answer
enot [183]

Answer:

It should be correct.

8 0
3 years ago
A car valued at £18000 at the start of 2017, depreciated in value by 5% each year for 3 years. How much did it lose in value ove
Anna11 [10]

<u>Answer:</u>

The amount lost over the 3 years s 2567.25£  

<u>Explanation:</u>

$\mathrm{F}=\mathrm{I} \times\left(1-\left(\frac{r}{100}\right)\right)^{\mathrm{n}}$

where F = final value after n years

I = initial value of the car in 2017 = £18000 (given)

Since the value is depreciated 5% every year for 3 years,

r = percentage rate of depreciation = 5% (given)

n = 3 years

Substituting these values in formula, we get

$\mathrm{F}=18000 \times\left(1-\frac{5}{100}\right)^{3}$

= $18000 \times\left(\frac{95}{100}\right)^{3}$

 = 15432.75£ which is the value of the car after 3 years

Finally 18000-15432.75 = 2567.25£ is the amount lost over this period.  

6 0
3 years ago
Find the solutions of the quadratic equation 14x^2+9x+10=014x
Vesnalui [34]

Answer:

Option B. x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

14x^{2}+9x+10=0

so

a=14\\b=9\\c=10

substitute in the formula

x=\frac{-9(+/-)\sqrt{9^{2}-4(14)(10)}} {2(14)}

x=\frac{-9(+/-)\sqrt{-479}} {28}

Remember that

i=\sqrt{-1}

substitute

x=\frac{-9(+/-)\sqrt{479}i} {28}  

x=-\frac{9}{28}(+/-)\frac{\sqrt{479}}{28}i

5 0
3 years ago
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