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3 years ago

Which number below eaqul 129000

Mathematics

1 answer:

katen-ka-za [31]3 years ago

5 0

Answer: There are no numbers shown

Step-by-step explanation:

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A building casts a shadow 40 ft long when the sun's angle of elevation is 58 degrees. find the height of the building

zzz [600]

The building is 64 ft tall

3 0

3 years ago

Convert the quadratic equation, <br>f (x) = 2(x - 3)2 +1, into standard form.

laiz [17]

Step-by-step explanation:

........4x-y=11........

8 0

3 years ago

Other term to write 2x+6

eimsori [14]

You could factor it as 2(x + 3)

7 0

3 years ago

Work out the volume of the shape

pashok25 [27]

Answer:

$\large\boxed{V=\dfrac{1,421\pi}{3}\ cm^3}$

Step-by-step explanation:

We have the cone and the half-sphere.

The formula of a volume of a cone:

$V_c=\dfrac{1}{3}\pi r^2H$

r - radius

H - height

We have r = 7cm and H = (22-7)cm=15cm. Substitute:

$V_c=\dfrac{1}{3}\pi(7^2)(15)=\dfrac{1}{3}\pi(49)(15)=\dfrac{735\pi}{3}\ cm^3$

The formula of a volume of a sphere:

$V_s=\dfrac{4}{3}\pi R^3$

R - radius

Therefore the formula of a volume of a half-sphere:

$V_{hs}=\dfrac{1}{2}\cdot\dfrac{4}{3}\pi R^3=\dfrac{2}{3}\pi R^3$

We have R = 7cm. Substitute:

$V_{hs}=\dfrac{2}{3}\pi(7^3)=\dfrac{2}{3}\pi(343)=\dfrac{686\pi}{3}\ cm^3$

The volume of the given shape:

$V=V_c+V_{hs}$

Substitute:

$V=\dfrac{735\pi}{3}+\dfrac{686\pi}{3}=\dfrac{1,421\pi}{3}\ cm^3$

7 0

3 years ago

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0

4 years ago