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Natasha2012 [34]

4 years ago

9

in the first 1/2 mile of the trail, Jayden posts 3 signs that describe local birds. If the signs are equal distance apart , What

is the distance between the second and third sign?

1 answer:

My name is Ann [436]4 years ago

8 0

1/2 of a mile is divided equally into 3 parts- we need to find the distance between the second and third sign.

1/2 x 1/3 is 1/6.

This means that each sign is 1/6th of a mile apart.

If you have further confusions about the question, please feel free to message me and I will help to the best of my ability!

Hope this Helps!
-Sinnamin

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What's 76 divided by 9 equals something or 9 / 76 equals something what's the answer easy one

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The answer is 76 divided by 9 equals 8.44

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If c=25 and B=75 degrees, find a. Round to the nearest tenth.

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Answer:

a=6.9

Step-by-step explanation:

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4 years ago

Please help! While mining, Mike found a large metal bar that contained 6 ounces of lead. Mike also determined that the bar was 4

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Answer:

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Step-by-step explanation:

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3 years ago

Activity 4: Performance Task

Nookie1986 [14]

An arithmetic progression is simply a progression with a common difference among consecutive terms.

<em>The sum of multiplies of 6 between 8 and 70 is 390</em>
<em>The sum of multiplies of 5 between 12 and 92 is 840</em>
<em>The sum of multiplies of 3 between 1 and 50 is 408</em>
<em>The sum of multiplies of 11 between 10 and 122 is 726</em>
<em>The sum of multiplies of 9 between 25 and 100 is 567</em>
<em>The sum of the first 20 terms is 630</em>
<em>The sum of the first 15 terms is 480</em>
<em>The sum of the first 32 terms is 3136</em>
<em>The sum of the first 27 terms is -486</em>
<em>The sum of the first 51 terms is 2193</em>

<em />

<u>(a) Sum of multiples of 6, between 8 and 70</u>

There are 10 multiples of 6 between 8 and 70, and the first of them is 12.

This means that:

$\mathbf{a = 12}$

$\mathbf{n = 10}$

$\mathbf{d = 6}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{10} = \frac{10}2(2*12 + (10 - 1)6)}$

$\mathbf{S_{10} = 390}$

<u>(b) Multiples of 5 between 12 and 92</u>

There are 16 multiples of 5 between 12 and 92, and the first of them is 15.

This means that:

$\mathbf{a = 15}$

$\mathbf{n = 16}$

$\mathbf{d = 5}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*15 + (16 - 1)5)}$

$\mathbf{S_{16} = 840}$

<u>(c) Multiples of 3 between 1 and 50</u>

There are 16 multiples of 3 between 1 and 50, and the first of them is 3.

This means that:

$\mathbf{a = 3}$

$\mathbf{n = 16}$

$\mathbf{d = 3}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{16}2(2*3 + (16 - 1)3)}$

$\mathbf{S_{16} = 408}$

<u>(d) Multiples of 11 between 10 and 122</u>

There are 11 multiples of 11 between 10 and 122, and the first of them is 11.

This means that:

$\mathbf{a = 11}$

$\mathbf{n = 11}$

$\mathbf{d = 11}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{16} = \frac{11}2(2*11 + (11 - 1)11)}$

$\mathbf{S_{11} = 726}$

<u />

<u>(e) Multiples of 9 between 25 and 100</u>

There are 9 multiples of 9 between 25 and 100, and the first of them is 27.

This means that:

$\mathbf{a = 27}$

$\mathbf{n = 9}$

$\mathbf{d = 9}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{9} = \frac{9}2(2*27 + (9 - 1)9)}$

$\mathbf{S_{9} = 567}$

<u>(f) Sum of first 20 terms</u>

The given parameters are:

$\mathbf{a = 3}$

$\mathbf{d = 3}$

$\mathbf{n = 20}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{20} = \frac{20}2(2*3 + (20 - 1)3)}$

$\mathbf{S_{20} = 630}$

<u>(f) Sum of first 15 terms</u>

The given parameters are:

$\mathbf{a = 4}$

$\mathbf{d = 4}$

$\mathbf{n = 15}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{15} = \frac{15}2(2*4 + (15 - 1)4)}$

$\mathbf{S_{15} = 480}$

<u>(g) Sum of first 32 terms</u>

The given parameters are:

$\mathbf{a = 5}$

$\mathbf{d = 6}$

$\mathbf{n = 32}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{32} = \frac{32}2(2*5 + (32 - 1)6)}$

$\mathbf{S_{32} = 3136}$

<u>(g) Sum of first 27 terms</u>

The given parameters are:

$\mathbf{a = 8}$

$\mathbf{d = -2}$

$\mathbf{n = 27}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{27} = \frac{27}2(2*8 + (27 - 1)*-2)}$

$\mathbf{S_{27} = -486}$

<u>(h) Sum of first 51 terms</u>

The given parameters are:

$\mathbf{a = -7}$

$\mathbf{d = 2}$

$\mathbf{n = 51}$

The sum of n terms of an AP is:

$\mathbf{S_n = \frac n2(2a + (n - 1)d)}$

Substitute known values

$\mathbf{S_{51} = \frac{51}2(2*-7 + (51 - 1)*2)}$

$\mathbf{S_{51} = 2193}$

Read more about arithmetic progressions at:

brainly.com/question/13989292

4 0

2 years ago

Read 2 more answers

-11y = 6(2+1) - 13y

iris [78.8K]

Answer:

12

Step-by-step explanation:

Let's put the equations in standard form. For the first equation, we have:

−11y=6(z+1)-13y

2y−6z=6

y−3z=3

The second equation is:

4y−24=c(z−1)

4y−cz=24−c

If we multiply the first equation by 4, we get:

4y-12z=12

Comparing the two equations, we see that if c=12, both equations will be the same and there will be infinitely many solutions.

The correct value of c is 12.

7 0

3 years ago