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2 years ago

5. The population of Springfield has increased 33.4% since last year.-

Mathematics

1 answer:

lutik1710 [3]2 years ago

6 0

Answer:

33.4%= 33.4 ÷100×9000= 3006.0

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If tan theta =3, find the value of tan theta + tan (theta+pi) + tan (theta +2pi)

sweet-ann [11.9K]

Answer:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

Step-by-step explanation:

Given

$tan\theta = 3$

Required

Find

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$

Calculate $tan(\theta + \pi) \ and\ tan(\theta+2\pi)$

Using tan rule

$tan(\theta + \pi) = \frac{tan\theta + tan\pi}{1 - tan\theta tan\pi}$

So:

$tan(\theta + \pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + \pi) = \frac{tan\theta}{1 }$

$tan(\theta + \pi) = \tan\theta$

$tan(\theta+2\pi)$

$tan(\theta + 2\pi) = \frac{tan\theta + tan2\pi}{1 - tan\theta tan2\pi}$

$tan(\theta + 2\pi) = \frac{tan\theta + 0}{1 - tan\theta *0}$

$tan(\theta + 2\pi) = \tan\theta$ '

'So:

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi)$ $= tan\theta +tan\theta +tan\theta$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 3+3+3$

$tan\theta + tan(\theta + \pi) + tan(\theta+2\pi) = 9$

4 0

2 years ago

Find the solution to the initial value problem (1 + x^{5}) y' + 5 x^{4} y = 5 x^{8}

natali 33 [55]

$(1+x^5)y'+5x^4y=5x^8$
$\bigg((1+x^5)y\bigg)'=5x^8$
$(1+x^5)y=5\displaystyle\int x^8\,\mathrm dx$
$(1+x^5)y=\dfrac59x^9+C$
$y=\dfrac{5x^9+C}{9(1+x^5)}$

8 0

3 years ago

Can you please help me with this

dem82 [27]

The answer is 63

Glad I could help :)

6 0

2 years ago

Read 2 more answers

What percent is equal to the decimal 0.13? A. 0.013% B. 1.3% C. 13% D. 130%

mario62 [17]

0.13 <== the decimal is in the tenth place.

We could also say it is equal to

13/100

Or,

"C" 13%

I hope this helps!
~kaikers

3 0

3 years ago

Read 2 more answers

What are the vertical and horizontal asymptotes for the function f(x)= 3x2/x2-4

Alecsey [184]

Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: $f(x)=\dfrac{3x^2}{x^2-4}$

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. $x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2$

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at $y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}$

i.e. $y=\dfrac{3}{1}=3$

Hence, f(x) will have horizontal asymptote at y=3.

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3 years ago