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antiseptic1488 [7]
2 years ago
14

Find the LCM for 5,9

Mathematics
1 answer:
Alex2 years ago
5 0

Answer:

45 :)

Step-by-step explanation:

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Let h = for 0   0.01 m, and h = for   0.01 m. (a) find j everywhere. (b) what is j at  = 0? (c) is there a filamentary cu
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4 0
3 years ago
A scientist wants to find the radius, in feet, of this hemispherical dome. He found that the surface area of the entire sphere c
skelet666 [1.2K]
The surface area of a sphere by definition is given by:
 A = 4 * π * r ^ 2
 Where,
 r: radius of the sphere.
 Substituting the values we have:
 294 = 4 * π * r ^ 2
 We cleared the radio:
 r ^ 2 = (294) / (4 * π)
 Answer:
 An equation that I could use to find the dome's radius is:
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7 0
3 years ago
The bases of a right prism are rhombi with area A = 44 cm2. The height of the prism is h=2.3 dm. Find the volume V of the prism.
Allushta [10]
The volume is 1012 cm³.

The volume of a right prism is found by multiplying the area of the base and the height.  First we need to convert the height to centimeters.

1 dm = 10 cm
2.3 dm = 2.3(10) = 23 cm

Now we have 44(23) = 1012 cm³
8 0
3 years ago
Please answer this question only if you know the answer! 30 points and brainliest!
Allisa [31]

Answer:

(4, 3)

Step-by-step explanation:

If this point is reflected about the x-axis, the x-coordinate does not change.  The original y-coordiate, -3, becomes +3.

A': (4, 3)

5 0
3 years ago
Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet
zmey [24]

Answer:

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

hypotenuse^2=perpendicular^2+base^2

hypotenuse^2=(5)^2+(12)^2

hypotenuse^2=25+144

hypotenuse^2=169

Taking square root on both sides.

hypotenuse=13

In a right angled triangle

\sin \theta = \dfrac{opposite}{hypotenuse}

\sin \theta = \dfrac{5}{13}

\sec \theta = \dfrac{hypotenuse}{adjacent}

\sec \theta = \dfrac{13}{12}

In second quadrant only sine and cosecant are positive.

\sin \theta = \dfrac{5}{13} and \sec \theta = -\dfrac{13}{12}

6 0
3 years ago
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