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Alchen [17]
3 years ago
6

243 as a power of 3

Mathematics
2 answers:
nekit [7.7K]3 years ago
7 0

Answer:

3^5

plz mark brainliest✌️✌️

Fudgin [204]3 years ago
4 0

Answer:

243 as a power of 3

= 3^5

=243

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Y+1=3/4(x+3) in general form pls??
Vikentia [17]

Answer:

Step-by-step explanation:

y+1 =(3/4)x+9/4

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

7 0
3 years ago
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4/5 as a whole number plz :)
aev [14]
The answer is 1 because4 is close to 5
8 0
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The range in temperature for the month of November in El Paso,Texas is 44 degrees. If the record high temperature is 92°F then w
gizmo_the_mogwai [7]

Answer:

33 degrees is the low temp

Step-by-step explanation:

4 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<span>
b. Incenter: the intersection of bisector lines
c. Centroid: </span>the intersection of median lines<span>
d. Orthocenter: </span>the intersection of altitude lines
5 0
3 years ago
I need help on numbers 11-16
agasfer [191]
Recall your SOH CAH TOA, or \bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad \qquad &#10;% cosine&#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;&#10;\\ \quad \\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}

now, bear in mind that, the opposite side, to an angle, is the side right "in front of it", that is, if you were to put your eye on that angle, "the wall" you'd see on the other end, is the opposite side

the adjacent side, adjacent = next to, is the side that's touching the angle itself

and the hypotenuse, is always the slanted and longest side of all three

for example on 16, tangent of Z

if you put one eye on Z, you'll see on the other end, the side of 30 units
the adjacent side is the one touching Z, or the 40 units side

\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(Z)=\cfrac{30}{40}&#10;\\\\\\&#10;\textit{which can be simplified to }\cfrac{3}{4}&#10;

and you'd do all others, the same way, using those ratios

use \bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad \qquad &#10;% cosine&#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;&#10;\\ \quad \\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}
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3 years ago
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