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telo118 [61]
2 years ago
11

Someone please help me!!! ASAP

Mathematics
2 answers:
inn [45]2 years ago
6 0
102 3/5 steps


The bridge is 307 4/5 m long
Her steps are 3/10 m long
Divide the length of the bridge by the length of her steps

First change the 307 4/5 to 8/10 so we have a common denominator
(307 8/10)/(3/10)
Next change 307 8/10 to an improper fraction
Multiply the whole number by the denominator than add the numerator
(3078/10)/(3/10)=1,026/10
Now divide the numerator by the denominator
1,026/10=102 6/10
Which simplifies to 102 3/5
melamori03 [73]2 years ago
6 0
102 3/5 is how many steps
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Express each logarithm in terms of In 2 and In 5.
Sladkaya [172]

Answer:

5ln2-3ln5

2ln5-ln2

Step-by-step explanation:

∵ 32=2^{5}

∵ 125=5^{3}

∵ ln\frac{32}{125}=ln\frac{2^{5}}{5^{3}}

∴ ln2^{5}-ln5^{3}=5ln2-3ln5

∵ 12.5=\frac{125}{10}=\frac{25}{2}=\frac{5^{2}}{2}

∴ ln12.5=ln\frac{5^{2}}{2}

∴ ln5^{2}-ln2=2ln5-ln2

3 0
3 years ago
Solve the one-step equation <br> -9 = u/-6
sladkih [1.3K]

Answer:

3

Step-by-step explanation:

-6 x 3 = -9

4 0
2 years ago
A track-and-field athlete releases a javelin. The height of the javelin as a function of time is shown on the graph below. Use t
Georgia [21]
Part 1:
 
 For this case we must see in the graph the axis of symmetry of the given parabola.
 We have then that the axis of symmetry is the vertical line t = 2.
 Answer:
 
The height of the javelin above the ground is symmetric about the line t = 2 seconds:

 
Part 2:

 
For this case, we must see the time t for which the javelin reaches a height of 20 feet for the first time.
 We then have that when evaluating t = 1, the function is h (1) = 20. To do this, just look at the graph.
 Then, we must observe the moment when it returns to be 20 feet above the ground.
 For this, observing the graph we see that:
 h (3) = 20 feet
 Therefore, a height of 20 feet is again reached in 3 seconds.
 Answer:
 
The javelin is 20 feet above the ground for the first time at t = 1 second and again at t = 3 seconds
5 0
3 years ago
Write down a fraction whose value lies between 3 and 4; whose denominator is a multiple of 5; whose numerator is a multiple of 1
slamgirl [31]

Answer:

  33/10 and 55/15

Step-by-step explanation:

Possible denominators that are multiples of 5 less than 17 are 5, 10, 15.

Corresponding numerator ranges are [15, 20], [30, 40], and [45, 60]. In only two of these ranges are there any multiples of 11.

  33 in [30, 40]

  55 in [45, 60]

So, there are only two possible fractions meeting your requirement:

  33/10 and 55/15

6 0
3 years ago
The sum of the squares of three consecutive integers is 509. Determine the integers.
GalinKa [24]

Answer:

12, 13, 14

Step-by-step explanation:

Denote the integers as:

x

x+1

x+2

The sum of their squares, so that would be;

(x^(2)) + (( x + 1 )^(2)) + (( x + 2 )^(2)) = 509

write out the squares

x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 509

combine like terms

3x^2 + 6x + 5 = 509

inverse operations

3x^2 + 6x + 5 = 509

                -5     -5

3x^2 + 6x = 504

factor

3x^2 + 6x = 504

3 ( x^2 + 2x ) =504

Inverse operations

3 ( x^2 + 2x ) = 504

/3                       /3

x^2 + 2x = 168

Factor again

x ( x + 2 ) = 168

At this point, it should be obvious that x is 12 (because 12 * 14 = 168)

So now substitute back into the consecutive numbers

x = 12

x + 1 = 13

x + 2 = 14

6 0
3 years ago
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