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3 years ago

14

The value of the digit 6 in 36,408 is ______times the value of 6 in 27,364

1 answer:

Ne4ueva [31]3 years ago

4 0

100 times bigger then the other 6

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Solve the equation.<br> 5/3p+2/3=5+p/2p<br> a. p=12<br> b. p= 5<br> c. p= 9<br> d. p=6

Kobotan [32]

The answer is B) p=5

Steps :
1) Simplify : 5 + p / 2p
2) simplify : 2/3
3) simplify : giving u 5/3p

If you need this problem worked out let me know

6 0

4 years ago

Read 2 more answers

c2*sin(wt)+c1*cos(wt)= A*Sin(wt+phi), where c2=Acos(Phi) and c1=Asin(Phi). They ask me to find the amplitude of the function 2 s

Iteru [2.4K]

The amplitude of $A\sin(\omega t+\phi)$ is the absolute value of $A$ . So first you need to condense the given function into one sine expression.

Recall that

$A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi$

so you need to choose $\phi$ and $\omega$ accordingly.

If we line up the terms of the given function with the expanded one above, we should have

$2\sin(4\pi t)+5\cos(4\pi t)\implies\begin{cases}A\cos\phi=2\\A\sin\phi=5\\\omega=4\pi\end{cases}$

Now, using the Pythagorean identity,

$(A\sin\phi)^2+(A\cos\phi)^2=2^2+5^2\implies A^2=29\implies A=\pm\sqrt{29}$

so the amplitude is √29.

Just for completeness, we also get

$\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)+n\pi$

where $n$ is any integer.

7 0

3 years ago

12. a) Solve the following|x-1|<2.

Aleksandr-060686 [28]

<h3>Answer: -1 < x < 3</h3>

Work Shown:

| x-1 | < 2

-2 < x-1 < 2 ... see note1 below

-2+1 < x-1+1 < 2+1 ... see note2

-1 < x < 3

Note1: use the rule that |x| < k turns into -k < x < k for some positive real number k.
Note2: Add 1 to all sides to isolate x

3 0

2 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.

5 0

3 years ago

Pls give me the answer....

worty [1.4K]

7½ – 4/7 ×7/8

= 15/2 – 1/2

= 7

3 0

3 years ago

Read 2 more answers