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2 years ago

Write 145,567 in expanded notation

Mathematics

1 answer:

kakasveta [241]2 years ago

8 0

Answer:

100000+45000+500+60+7

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How many solutions are there to the equation 10= a + b + c with each of a, b, and c a whole number greater than or equal to zero

djverab [1.8K]

If a is fixed and b,c are unknowns then the equation b+c=10-a has 11-a solutions. They are pairs (b,c): (0,10-a), (1,9-a), (2,8-a), ... (10-a,0). As a runs from 0 to 10 we have total number of solutions (11-0)+(11-1)+...(11-1)=11+10+...+1=(1+11)*11/2=66.

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3 years ago

An item was sold at a loss of 20%. Had it been sold for Rs 3000more then the profit would have been 10%. What is the cost price

vagabundo [1.1K]

Answer:

Cost Price = Rs 10000

Step-by-step explanation:

Assume:

Cost of the item = x

Item was sold at a loss of 20%:

Loss = 20% of x = 0.2x

Item sold = x - 0.2x = 0.8x

Item sold at a profit of 10%:

Profit = 10% of x = 0.1x

item sold = x + 0.1x = 1.1x

Solve:

Difference = 1.1x - 0.8x = 0.3x

0.3x = Rs 3000

x = Rs 3000 ÷ 0.3

x = Rs 10000

3 0

2 years ago

Suppose f varies directly as g, and f varies inversely as h. Find g when f = 10 and h = –12, if g = 56 when h = –2 and f = –7. R

kow [346]

Answer:

-480

Step-by-step explanation:

If f varies directly as g, and inversely as h, then we can write the variation equation:

$f=\frac{kg}{h}$ , where k is the constant of variation.

If g=56 when h=-2 and f=-7, then we substitute these values into the formula to find the value of k.

$-7=\frac{k(56)}{-2}$

$-7\times -2=56k$

$k=\frac{14}{56}=\frac{1}{4}$

The equation now becomes:

$f=\frac{g}{4h}$

if f=10 and h=-12, then;

$10=\frac{g}{-48}$

$g=-48\times 10=-480$

The correct answer is B.

4 0

2 years ago

Learning Task 1: Find the value of th

shepuryov [24]

Answer:

a=5your answer right answer right answer

8 0

2 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

2 years ago

Write 145,567 in expanded notation​

Write 145,567 in expanded notation