Michael threw for 1,654 yards in his first five games. At this rate, how many yards will he have thrown for in fifteen games? ✗
2 answers:
Answer:
4,962 yards.
Step-by-step explanation:
We are given that,
For the first 5 games, Michael threw 1,654 yards.
Since, we see that, the relationship between the yards and the games are directly proportional.
Then for 15 games,
Michael will have thrown for yards i.e. 1654 × 3 = 4,962 yards
Hence, for 15 games, Michael have thrown for 4,962 yards
Answer: 4,962 yards
Step-by-step explanation:
Given: Michael threw for 1,654 yards in his first five games.
The rate of threw =
Then , The rate of threw = yard/ minute.
i.e. the distance of yards thrown in 1 minute = yard
If the rate remains same , then the distance of yards thrown for in fifteen games is given by :-
Hence, At this rate, 4,962 yards will he have thrown for in fifteen games.
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Answer:
the formula for a square pyramid is so I THINK its B please do not base ur answer from my opinion
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
To make it easy let's start by organizing our information :
- AC=12 AND BD=8
- ABCD is a rhombus
- K and L are the midpoints of sides AD and CD
- we notice that the rhombus ABCD is divided into four right triangles
What do you think of when you hear a right triangle ?
- The pythagorian theorem !
AC and BD are khown so let's focus on them .
If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?
Simply because they are the diagonals of a rhombus .
ow let's apply the pythagorian theorem :
- (AC/2)² + (BD/2)² = BC²
- 6²+4²=52
- BC²= 52⇒
=BC
Now we khow that : AB=BC=CD=AD=
This isn't enough . Let's try to figure out a way to calculate the length of KL wich is the base of the triangle
- KL is parallel to AC
- k is the midpoint of AD and L of DC
I smell something . yes! Thales theorem
- KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
- KL/12=
- KL/12=1/2⇒ KL=6
Now we have the length of the base kl
Now the big boss the height :
- notice that you khow the length of KL
- BD crosses kl from its midpoint and DL =
What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D
- DL²=(KL/2)²+D²
- 52/4= 9+ D²
- D² = 52/4-9 +4 SO D=2
now the height of the trigle is H= BD-D= 8-2=6
NOw the area of the triangle is :
- A=(KL*H)/2 ⇒ A= (6*6)/2=18
THE ANSWER IS 18 SQ.UN
The answer is
steps
#1) Try to leave a by itself by dividing any variable next to it
you can only divided these variables as all these variables in the original equation. are being multiplied. The opposite if multiplication is division.
steps
#1) Try to leave a by itself by dividing any variable next to it
you can only divided these variables as all these variables in the original equation. are being multiplied. The opposite if multiplication is division.