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yaroslaw [1]
4 years ago
12

Someone help with these two ??

Mathematics
1 answer:
kkurt [141]4 years ago
6 0
1) 0.20
2)0.58
Hope this helped
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129÷21 show work need help
scZoUnD [109]
I don’t think you need to continue any further but there you go

8 0
3 years ago
How do I do this again? please help ​
Mars2501 [29]

Answer:

six

Step-by-step explanation:

5 0
3 years ago
8 y ^ 6 raised to the power of 1/3
Mkey [24]
2y^2

multiply all exponents by 1/3 to get

8^1/3 y^6/3.

then that gives 2y^2
6 0
3 years ago
Find 'x' in the following proportion 4:8::x:16 ? 4 4 Express the following statement alcohrnically​
ipn [44]

Answer:

x= 8

Step-by-step explanation:

\frac{4}{8}=\frac{x}{16}

4*16=64

64/8=8

8 0
3 years ago
Someone help me please.
mrs_skeptik [129]

Step-by-step explanation:

f(x) = -0.01x² + 0.7x + 6.1

a) f(x) is a downward facing parabola, so the maximum height is at the vertex.  The vertex of a parabola can be found using x = -b/(2a).

x = -0.7 / (2 × -0.01)

x = 35

f(35) = -0.01(35)² + 0.7(35) + 6.1

f(35) = 18.35

The maximum height is 18.35 feet.

b) The maximum horizontal distance is when the ball lands, or when f(x) = 0.

0 = -0.01x² + 0.7x + 6.1

0 = x² − 70x − 610

Solve with quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

x = [ -(-70) ± √((-70)² − 4(1)(-610)) ] / 2(1)

x = (70 ± √7340) / 2

x = 35 ± √1835

x = -7.84, 77.84

x can't be negative, so x = 77.84.  The ball's maximum horizontal distance is 77.84 feet.

c) When the ball is first launched, x = 0.  The height at that position is:

f(0) = 6.1

The ball is launched from an initial height of 6.1 feet.

7 0
3 years ago
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