Plzzz help me I am timed plzzzzz

Nonamiya [84]

$16. \ \ V= \pi r^2h \ \to \ r^2= \cfrac{V}{ \pi h} \ \to \ \boxed{r= \sqrt{ \frac{V}{ \pi h} } } \\ \\ \\ 17.\ \ d= \frac{1}{2} at^2 \ \to \ 2d= at^2 \ \to \ t^2= \cfrac{2d}{a} \ \to \ \boxed{t= \sqrt{\frac{2d}{a}}} \\ \\ \\ 18. \ \ S=180(n-2) \ \to \ \cfrac{S}{180}=n-2 \ \to \ \boxed{n= \frac{S}{180}+2 }$

6 0

4 years ago

Parabolas. Please help me

olganol [36]

x² = 16 y
x = 0
(x + 4)² = 2/3 (y - 2)
Gayle identifies that the vertex of the parabola is (3, -1) . The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x = 0.
(y - 6)² = 4p (x - 3)

Step-by-step explanation:

1. First figure

We plot the parabola as given in the attached diagram.

As it is facing upwards, the equation goes as x² = 4py

where, p = 4 (refer the attached diagram)

x² = 4py

x² = 4 (4) y

∴, standard form of parabola is x² = 16 y

2. Second figure

(y + 3)² = 4 (x - 1)

Comparing the given equation with the standard form

(y - k)² = 4p (x - h)

Now from this equation we get to know that

h = 1

p = 1

Directrix is x = (h - p)

So, x = 0

3. Third figure

3x² + 24x - 2y + 52 = 0

3x² + 24x = 2y - 52

3 (x² + 8x) = 2 (y - 26)

(x² + 8x) = 2 (y - 26) / 3

Adding 16 on both sides,

x² + 8x + 16 = 2 (y - 26) / 3 + 16

(x + 4)² = 2/3 y - 52/3 + 16

(x + 4)² = 2/3 y - 4/3

(x + 4)² = 2/3 (y - 2)

4. Fourth figure

(y + 1)² = 12 (x - 3)

Comparing the given equation with the standard form

(y - k)² = 4p (x - h)

Now from this equation we get to know that

k = -1

h = 3

p = 3

Gayle identifies that the vertex of the parabola is (3, -1) . The parabola opens right, and the focus is 3 units away from the vertex. The directrix is 6 units from the focus. The focus is the point (6, -1). The directrix of the equation is x = 0.

5. Fifth figure

Focus = (2, 6)

Directrix is x = 4

Therefore, it follows the standard form

(y - k)² = 4p (x - h)

Directrix is given by x = h-p = 4

Focus is given by (h + p, k) = (2, 6)

Solving for (h - p) = 4, (h + p) = 2

2 - p - p = 4

-2p = 2

p = -1

Hence, h = 3

Therefore, the standard form can be written as

(y - 6)² = 4p (x - 3)

7 0

3 years ago

Confused can someone help

Alex

Answer:

B. 3x^2 - 2x - 1.

Step-by-step explanation:

(f + g)x = f(x) + g(x)

= -4x + 3 + 3x^2 + 2x - 4

Adding like terms we get the answer:

= 3x^2 - 2x - 1.

7 0

4 years ago

Read 2 more answers

Alia has swimming class every fourth day and diving class every sixth day. If she had swimming class and diving class on 5 june,

dolphi86 [110]

Answer:

June 17

Step-by-step explanation:

the least common multiple of 4 and 6 is 12. Add 12 to June 5 and you get June 17

3 0

3 years ago

1. M is the midpoint of LN and O is the midpoint of NP.

Reptile [31]

1. M is the midpoint of LN and O is the midpoint of NP. This makes the triangle MNO equal to half of LNP. Then you can get this equation
MO= (1/2) LP

If you insert MO = 2x +6 and LP = 8x – 20 the calculation would be:
2x+6= (1/2)( 8x-20)
2x+6= 4x-10
2x-4x= -10 - 6
-2x= -16
x=8

2. Centroid is the point that intersects with three median lines of the triangle. The centroid should divide the median lines into 1:2 ratio. In AC lines, A located in the base so A.F:FC would be 1:2

Then, the answer would be:
A.F= 1/(1+2) * AC
A.F= 1/3 * 12= 4

FC= 2/(1+2) * AC
FC= 2/3 * 12= 8

3. Since
∠BAD=∠DAC
∠ABD=∠ACD
AD=AD
The triangle ABD and ACD are similar. You can get this equation
BD=DC
x+8= 3x+12
x-3x= 12-8
-2x=4
x=-2

DC=3x+12= 3(-2) +12= 6

4. Orthocenter made by intersection of triangle altitude
A
BC lines slope would be (-4)-(-1)/1-4= -3/-3= 1. The altitude line slope would be -1, the function would be:
y=-x +a
0= 1+a
a=-1
y=-x-1
B
AC lines slope would be (-4)-(-1)/1-0= -3. The altitude line slope would be 1/3, the function would be:
y=1/3x+a
-1=1/3(4)+a
a=-7/3
y=1/3x - 7/3

C
BC lines slope would be (-1)-(-1)/4 = 0/4.
The line would be
0=x+a
a=-1
0=x-1
x=1

y=-x-1 = 1/3x-7/3
-x-(1/3x)=-7/3 +1
-4/3x= -4/3
x=1

y=-x-1
y=-1-1= -2
The orthocenter would be (1,-2)

5.
a. Circumcenter: the intersection of perpendicular bisector lines
b. Incenter: the intersection of bisector lines
c. Centroid: the intersection of median lines
d. Orthocenter: the intersection of altitude lines

5 0

3 years ago