Answer:
The number of times Ellis get to bat is 558.
Step-by-step explanation:
Here, let us assume the number of times Ellis bat = m
So, the number of times Dwight bat = 17 times fewer than Ellis
= m - 17
Also, number of times Wade got to bat = 10 more times than Dwight
= ( m- 17 )+ 10
Total number of bat times = 1650
So, the number of times ( Wade + Dwight + Ellis) bat together = 1650
⇒ ( m- 17 )+ 10 + (m - 17) + m = 1650
or, 3 m - 34 + 10 = 1650
or, 3 m = 1674
⇒ m = 1674/3 = 558 , or m = 558
Hence, the number of times Ellis get to bat = m = 558.
Answer:
21-7x
Step-by-step explanation: gather like terms
25-4=21 -9x+2x=-7x
The distance ( S ) is represented by the equation:
S ( t ) = - 16 t² + 25
- 16 t² + 25 = 0
- 16 t² = - 25 / · ( - 1 )
16 t² = 25
t ² = 25 / 16
t = √( 25/16) = 5/4 = 1.25 s ( we will not accept the negative value )
Answer:
The acorn is moving trough the air for 1.25 seconds.
Answers:
u = 18
v = 20
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Work Shown:
The leg YZ is congruent to the left FH
This means YZ = FH and that leads to u+v-18 = 10u-8v
The hypotenuse ZX is congruent to the hypotenuse HG
This means ZX = HG and we get the equation 14v-14u+32 = v+u+22
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The system of equations we have so far is
u+v-18 = 10u-8v
14v-14u+32 = v+u+22
Let's get the equations into standard form
Start with the first equation
u+v-18 = 10u-8v
u+v = 10u-8v+18
u+v-10u+8v = 18
-9u+9v = 18
-9(u-v) = 18
u-v = 18/(-9)
u-v = -2
Note how we can solve for the variable u to get
u = v-2
which we'll use later
----------------------
Let's get the other equation into standard form as well
14v-14u+32 = v+u+22
14v-14u-v-u = 22-32
-15u+13v = -10
Now plug in u = v-2 and solve for v
-15u+13v = -10
-15(v-2) + 13v = -10
-15v+30+13v = -10
-2v+30 = -10
-2v = -10-30
-2v = -40
v = -40/(-2)
v = 20
Which means,
u = v-2
u = 20-2
u = 18
