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3 years ago

Work out the value of: (√3)^2

Mathematics

1 answer:

Viktor [21]3 years ago

7 0

Answer:

Step-by-step explanation:

Note that

( $\sqrt{a}$ )² = a, for example

( $\sqrt{100}$ )² = 10² = 100 ← value inside the radical

Thus

( $\sqrt{3}$ )² = 3

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Solve this equation 6(4.5y-12)=9

irinina [24]

6(4.5y - 12) = 9

(4.5y - 12) = 9/6

4.5y - 12 = 1.5

4.5y = 1.5 + 12

4.5y = 13.5

y = 13.5/4.5

y = 3

6 0

3 years ago

When records were first kept (t=0), the population of a rural town was 260 people. During the following years, the population gr

Firdavs [7]

Answer:

(a) The population after 15 years is 2678.

(b)Therefore the population P(t) at any time t>0 is

$P(t)= 45t+30 {t^{\frac32}}+260$

Step-by-step explanation:

Given that,

The population grew at a rate of

$P'(t)=45(1+\sqrt t)$

Integrating both sides

$\int P'(t) dt=\int 45(1+\sqrt t)dt$

$\Rightarrow \int P'(t) dt=\int (45+45\sqrt t)dt$

$\Rightarrow \int P'(t) dt=\int 45\ dt+\int 45\sqrt t\ dt$

$\Rightarrow P(t)= 45t+45\ \frac{t^{\frac12+1}}{\frac12+1}+c$ [ c is integration constant]

$\Rightarrow P(t)= 45t+45\ \frac{t^{\frac32}}{\frac32}+c$

$\Rightarrow P(t)= 45t+45\times\frac 23 \times {t^{\frac32}}+c$

$\Rightarrow P(t)= 45t+30 {t^{\frac32}}+c$

When t=0 , P(0)= 260

$\therefore 260= 45\times0+30\times {0^{\frac32}}+c$

$\Rightarrow c=260$

$\therefore P(t)= 45t+30 {t^{\frac32}}+260$

Therefore the population P(t) at any time t>0 is

$P(t)= 45t+30 {t^{\frac32}}+260$

To find the population after 15 years, we need to plug t=15 in the above expression.

$P(15)=( 45\times 15)+30( {15^{\frac32}})+260$

≈2678

The population after 15 years is 2678.

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3 years ago

Approximate the value of √44 to the nearest hundredth

mestny [16]

$2 \sqrt{11}$

$6.63$

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Work out the value of: (√3)^2​

Work out the value of: (√3)^2