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Sergio [31]
4 years ago
6

Jessica walks 12 miles in 3hrs. How many miles does she walk in one hour

Mathematics
1 answer:
ehidna [41]4 years ago
5 0
3 hours = 12 miles

1 hour = 12 ÷ 3 = 4 miles

Answer: 4 miles
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Lee has $1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he hav
Schach [20]

Answer:

Lee has 19 nickels and 8 dimes

Step-by-step explanation:

hello, I think I can help you with this

Step 1

Let

value of a nickel=$0.05

value of a dime=$0.10

number of nickels=N

number of nickels=D

According to the question data Lee has $1.75 in dimes and nickels.

0.05N+0.1D=1.75,equation(1)

also, the number of nickels is 11 more than the number of dimes.mathematical speaking

N=11+D,equation(2)

Step 2

replace N from equation (2) into equation (1)

0.05N+0.1D=1.75

0.05(11+D)+0.1D=1.75

Now, solve for D

0.55+0.05D+0.1D=1.75

0.15D=1.75-0.55

D=1.2/0.15

D=8

Step 3

replace the value of D into equation(2) to obtain N

N=11+8

N=11+8

N=19

Hence, Lee has 19 nickels and 8 dimes

Have a great day.

5 0
3 years ago
Help fast. do first question only and please show work thx.
amm1812
Y = -0.01x^2 + 0.5x + 3

when the height y = 8 
-0.01x^2 + 0.5x + 3 = 8
-0.01x^2 + 0.5x - 5 = 0

x =  13.82 and x = 36.18 so the ball is at least 8 ft high   from 13.82 and 36.18 feet inclusive.- from the player.

If the player is 30 ft from the net then the ball is likely to go over the net.
7 0
3 years ago
A newly discovered planet orbits a distant star with the same mass as the Sun at an average distance of 101 million kilometers.
Elden [556K]
Kepler's third law described the relation between semi-major axis (or average distance to the star) and the orbital period (how long it takes to complete one lap) as follows: 
                                           a^3 / p^2 = constant
In the case of our Solar system the constant is 1 
This means that, for this problem:
a^3 / p^2 = 1
p^2 = a^3
p = a^(3/2)
The semi major axis is given as 101 million km. We need to convert this into AU where 1 AU is approximately 150 million Km
101 million Km = (101x1) / 150 = 0.67 AU

Now, we substitute in the equation to get the orbital period as follows:
p = (0.67)^(3/2) = 0.548 earth years
3 0
3 years ago
PLEASE SOLVE AND CHECK. SHOW COMPLETE SOLUTION
vivado [14]

Answer:

t \leq 25

Step-by-step explanation:

(\sqrt{2t+1}) ^{2} + (\sqrt{2t-1}) ^{2} \leq  10^{2}

=> 2t +1 + 2t -1 \leq 100

=> 4t \leq 100

=> t \leq 100/4

=> t \leq 25

3 0
3 years ago
6th grade math help me pleaseeee
nekit [7.7K]

Answer:

10.1

<em>hope this helps!</em>

7 0
3 years ago
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