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zhuklara [117]
3 years ago
9

A cylinder has a diameter of 6 cm and a height of 14 cm. What is the approximate volume of a cone with the same diameter and hei

ght
as the cylinder?
Mathematics
1 answer:
Feliz [49]3 years ago
3 0

Answer:

volume of cone=1/3 ×π×(d/2)²h=1/3 ×22/7×6²/4×14=924cm³

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What is the slope of a straight line, which is perpendicular to the line y = 1.6x + 4?
e-lub [12.9K]
The answer is -0.625
8 0
2 years ago
A triangle is formed by the ordered pairs X(–5, 6), Y(7, 6), and Z(7, 2). If mX=18°, what is mZ?
Mamont248 [21]

Answer:

mZ is 72°

Step-by-step explanation:

The three points form a right triangle with a 90° angle at Y. Since a triangle has 180°, first subtract 180-90=90°. Next, subtract 90-18=72 because mX=18°. Therefore, mZ is 72°.

8 0
3 years ago
What is the volume of the composite figure?
sertanlavr [38]

Look at the picture.

We have 3 cuboids. The formula of a volume of a cuboid is:

V = Length × Width ×  Height = lwh

The cuboid 1 are congruent to the cuboid 2.

<h2>1&2</h2>

l = 9 in, w = 10 in, h = 10 in

V = (9)(10)(10) = 900 in³

<h2>3</h2>

l = 6 in, w = 10 in, h = 10 in

V = (6)(10)(10) = 600 in³

The volume of the composite prism:

V = 2(900) + 600 = 1800 + 600 = 2400 in³

8 0
3 years ago
What is the factorization of x²+8x+7?
Alika [10]
(x+1)(x+7)
= <span>x^2 + x +7x+7
= </span><span>x²+8x+7

answer
</span><span>D. (x+1)(x+7)</span>
5 0
3 years ago
Find the directional derivative of at the point (1, 3) in the direction toward the point (3, 1). g
Anika [276]

Complete Question:

Find the directional derivative of g(x,y) = x^2y^5at the point (1, 3) in the direction toward the point (3, 1)

Answer:

Directional derivative at point (1,3),  D_ug(1,3)  = \frac{162}{\sqrt{8} }

Step-by-step explanation:

Get g'_x and g'_y at the point (1, 3)

g(x,y) = x^2y^5

g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486

g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405

Let P =  (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\

|\bar{PQ}| = \sqrt{8}

The unit vector is therefore:

u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }

The directional derivative of g is given by the equation:

D_ug(1,3)  = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3)  = (486*\frac{2}{\sqrt{8} } ) +  (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3)  = (\frac{972}{\sqrt{8} } ) +  (\frac{-810}{\sqrt{8} } )\\D_ug(1,3)  = \frac{162}{\sqrt{8} }

8 0
3 years ago
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