7. Write >,<, or = in the circle to make

Point A is located in which quadrant? On a coordinate plane, point A is at (negative 3, 5). Quadrant I Quadrant II Quadrant III

Gennadij [26K]

Point a is located in Quadrant 1

5 0

3 years ago

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7 = x + 12<br><br><br><br> A. -19 B. -5 C. 4 D. 5

Elena L [17]

You have to get "X" by itself so you subtract 12, 7-12 is -5 so "X" = - 5

5 0

3 years ago

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Determine the value c so that each of the following

Levart [38]

This can help you

a) since it is discrete we need to think about the sum thing it it was continuous we would look at an integral thing so I think if I remember correctly we need to find c such that <span><span>∑<span>i=0</span>3</span>c(<span>x2</span>+4)=1
b) </span> problem b is a similar setup

4 0

4 years ago

From a circular cylinder of diameter 10 cm and height 12 cm are conical cavity of the same base radius and of the same height is

Nataliya [291]

<h3>Volume of the remaining solid = 628 cm^2</h3>

<h3>Whole surface area = 659.4 cm^2</h3>

Step-by-step explanation:

Now, Given that:-

Diameter (d) = 10 cm

So, Radius (r) = 10/2 = 5cm

Height of the cylinder = 12cm.

$volume \: of \: the \: cylinder \: = \pi {r}^{2} h$

$= > \pi \times {5}^{2} \times 12 {cm}^{3} = 300\pi {cm}^{3}$

Radius of the cone = 5 cm.

Height of the cone = 12 cm.

$slant \: height \: of \: the \: cone \: = \sqrt{ {h}^{2} + \: {r}^{2} }$

$= > \sqrt{ {5}^{2}+{12}^{2} } cm \: = 13cm$

Volume of the cone = 1/3 *πr^2h

$= > \frac{1}{3} \pi \times {5}^{2} \times 12 {cm}^{3} = 100\pi {cm}^{3}$

therefore, the volume of the remaining solid

$= 300\pi {cm}^{3} - 100\pi {cm}^{3} \\ = 200 \times 3.14 {cm}^{3} = 628 {cm}^{3}$

Curved surface of the cylinder =

$2\pi \: rh \: = 2\pi \times 5 \times 12 {cm}^{2} \\ = 120\pi {cm}^{2} .$

$curved \: surface \: of \: the \: cone \: = \pi \: rl \\ = \pi \times 5 \times 13 {cm}^{2} \\ = 65\pi {cm }^{2} \\ area \: of \: (upper)circular \: base \: \\ of \: cylinder \: = \\ = \pi \: {r}^{2} = \pi \times {5}^{2}$