Ask question

Ask question

All categories

3 years ago

9

PLS HELP ME ASAP FOR 6, and 7!!! (GIVING BRAINLIEST AND THANKS!!)

1 answer:

denis23 [38]3 years ago

3 0

For 6 the answer is c because 1 fourth goes into one and a half 6 times so you do 6 times 4 feet which is 24 feet

You might be interested in

A triangle with vertices A(0,0), B(6,0), and C(0,9) is rotated 360 about the y-axis. What is the volume of the figure?

Luden [163]

Given:

A triangle with vertices A(0,0), B(6,0), and C(0,9) is rotated 360 about the y-axis.

To find:

The volume of the figure.

Solution:

Points A(0,0) and C(0,9) lies on the y-axis. Length of AC is 9 units.

Points A(0,0) and B(6,0) lies on the x-axis. Length of AB is 6 units.

If a triangle with vertices A(0,0), B(6,0), and C(0,9) is rotated 360 about the y-axis, then it will form a cone with radius 6 units and height 9 units.

Volume of a cone is

$V=\dfrac{1}{3}\pi r^2h$

where, r is radius of the base and h is vertical height of the cone.

Putting r=6 and h=9, we get

$V=\dfrac{1}{3}\pi (6)^2(9)$

$V=\pi (36)(3)$

$V=108\pi$

Therefore, the volume of the figure is $108\pi$ sq. units.

7 0

3 years ago

On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.

9966 [12]

Answer:

The correct option is;

No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

Circle center (h, k) = (0, 0)

Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

$Distance \, formula = \sqrt{\left (x_{2}-x_{1} \right )^{2} + \left (y_{2}-y_{1} \right )^{2}}$

Distance² = (x - h)² + (y - k)² = r² which gives;

(3 - 0)² + (0 - 0)² = 3²

Hence r² = 3² and r = 3 units

We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

$\sqrt{\left (x_{2}-x_{1} \right )^{2} + \left (y_{2}-y_{1} \right )^{2}} = Distance$

Therefore;

(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²

$\sqrt{\left (2-0 \right )^{2} + \left (\sqrt{6} -0 \right )^{2}} = 10$

Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10

Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point (2, √6), does not lie on the circle.

4 0

3 years ago

Read 2 more answers

Please help I’m timed

kupik [55]

It’s the second one :)

3 0

3 years ago

Read 2 more answers

A wool sweater originally costs $42.00. if the sweater is on sale for 25% off, what is the new cost of the sweater? $

jolli1 [7]

42 (1- 25/100) = 63/2 = 31.5 dollars

answer is $31.50

hope it helps

8 0

3 years ago

Read 2 more answers

Can you answer these please many thanks

givi [52]

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$ .

To find: The expanded form of the given expressions.

(c) $4(m+n)=4m+4n$ (using $a(b+c)=ab+ac$ )

(d) $3(5-q)=15-3q$ (using $a(b-c)=ab-ac$ )

(e) $5(2c+1)=10c+5$ (using $a(b+c)=ab+ac$ )

(f) $3(2x-5)=6x-15$ (using $a(b-c)=ab-ac$ )

(g) $7(4b-1)=28b-7$ (using $a(b-c)=ab-ac$ )

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒ $3(2x+y-5)=3(2x+y)-15$ (using $a(b-c)=ab-ac$ )

⇒ $3(2x+y-5)=6x+3y-15$ (using $a(b+c)=ab+ac$ )

(i) $2(6a-4b+3)=2((6a-4b)+3)$

⇒ $2(6a-4b+3)=2(6a-4b)+6$ (using $a(b+c)=ab+ac$ )

⇒ $2(6a-4b+3)=12a-8b+6$ (using $a(b-c)=ab-ac$ )

(j) $6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$ (using $a(b+c)=ab+ac$ )

⇒ $6(m+n+p)=6m+6n+6p$ (using $a(b+c)=ab+ac$ )

(k) $y(y+2)=y^2+2y$ (using $a(b+c)=ab+ac$ )

(l) $g(g-3)=g^2-3g$ (using $a(b-c)=ab-ac$ )

(m) $n(4-n)=4n-n^2$ (using $a(b-c)=ab-ac$ )

(n) $a(b+c)=ab+ac$ (using $a(b+c)=ab+ac$ )

(o) $s(3s-4)=3s^2-4s$ (using $a(b-c)=ab-ac$ )

(p) $2x(x+5)=2x^2+10x$ (using $a(b+c)=ab+ac$ )

(q) $4y(x-3)=4xy-12y$ (using $a(b-c)=ab-ac$ )

(r) $5a(2b-5)=10ab-25a$ (using $a(b-c)=ab-ac$ )

(s) $4a(3b+2c)=12ab+8ac$ (using $a(b+c)=ab+ac$ )

(t) $5p(4p-5q)=20p^2-25pq$ (using $a(b-c)=ab-ac$ )

6 0

3 years ago