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olga55 [171]
3 years ago
12

F(x) = 8 – 3xEvaluate f(9)​

Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0
F(x) = 8 - 3x

f(9) = 8 - 3(9)
= 8 - 27
= -19

hope this helps!
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POLYGON BASICS (GEOMETRY) PLEASE HELP!!
Vesna [10]

Answer:

Polygons are named according to the number of sides and angles they have. The most familiar polygons are the triangle, the rectangle, and the square. A regular polygon is one that has equal sides. Polygons also have diagonals, which are segments that join two vertices and are not sides.



6 0
4 years ago
SOMEONE PLEASE HELP ME I WILLG IVE EVERYTHING!
Anvisha [2.4K]

Answer:

<h2>AB is around 33.18</h2><h2>BC is around 15.58 </h2>

Step-by-step explanation:

adjacent/hypotenuse is sine:

cos(28 degrees)=29.3/x

cos(28 degrees)*x=29.3

x=29.3/cos(28 degrees)

x=around 33.18

AB is around 33.18

opposite/adjacent is tangent

tan(28 degrees)=x/29.3

tan(28 degrees)*29.3=x

x=tan(28 degrees)*29.3

x=around 15.58

BC is around 15.58

4 0
3 years ago
The school that DeShawn goes to is selling tickets to the annual dance competition. On the first day of ticket sales, the school
coldgirl [10]

Answer:

Both child tickets and senior tickets cost $14.

Step-by-step explanation:

Since the school that DeShawn goes to is selling tickets to the annual dance competition, and on the first day of ticket sales, the school sold 10 senior citizen tickets and 8 child tickets for a total of $ 252, while the school took in $ 280 on the second day by selling 10 senior citizen tickets and 10 child tickets, to determine what is the price of each of one senior citizen ticket and one child ticket, the following calculation must be performed:

10 senior tickets + 8 child tickets = 252

10 senior tickets + 10 child tickets = 280

280 - 252 = 2 child tickets

28 = 2 child tickets

28/2 = 1 child ticket

14 = 1 child ticket

14 x 10 = 140

(280 - 140) / 10 = senior tickets

140/10 = 14 = senior tickets

Therefore, both child tickets and senior tickets cost $14.

3 0
3 years ago
Draw two polygons that have corresponding angles that are all congruent but are not similar to prove that angle congruence is no
Marina86 [1]

There are infinitely many ways to do this. One such way is to draw a very thin stretched out rectangle (say one that is very tall) and a square. Example: the rectangle is 100 by 2, while the square is 4 by 4.

Both the rectangle and the square have the same corresponding angle measures. All angles are 90 degrees.

However, the figures are not similar. You cannot scale the rectangle to have it line up with the square. The proportions of the sides do not lead to the same ratio

100/4 = 25

2/4 = 0.5

so 100/4 = 2/4 is not a true equation. This numerically proves the figures are not similar.

side note: if you are working with triangles, then all you need are two pairs of congruent corresponding angles. If you have more than three sides for the polygon, then you'll need to confirm the sides are in proportion along with the angles being congruent as well.

3 0
3 years ago
Help me, please with math
ArbitrLikvidat [17]

Answer:

y=16

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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