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3 years ago

7

Solve the ratio with solution nonsense=report1.23:672.67:89

1 answer:

aliya0001 [1]3 years ago

6 0

<h3>Step by step explanation:</h3><h3>23:67</h3><h3> 23÷67--Convert the ratio into a fraction using a÷b=a/b</h3><h3>23/67--Calculate the approximate value</h3><h3>0.343284--Round the number to the nearest ten thousand</h3><h3>0.3443--Write a decimal fraction</h3><h3>34.33/100--by definition p/100=p%</h3><h3>Answer:</h3><h2>1.34.33%</h2>

<h3>67:89</h3><h3> 67÷89--Convert the ratio into a fraction using a÷b=a/b</h3><h3>67/89--Calculate the approximate value</h3><h3>0.752809--Round the number to the nearest ten thousand </h3><h3>0.7528--Write as decimal fraction</h3><h3>75.28/100--by definition p/100=p%</h3><h3>Answer:</h3><h2>2.75.28%</h2>

<h2>That's the correct answer</h2><h2>Liked my answer </h2><h2>Brainliest</h2>

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Answer:

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Step-by-step explanation:

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True or false?is the point(16,4) included in this function: Y=2x or Y=x2 or y= x+12 or y=x-12

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Answer:

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Step-by-step explanation:

It is true because the graph of $y=x-12$ passes through the point $(16,4)$

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Substitute

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Therefore, the answer is true.

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3 years ago

There are 7 days in 1 week. How many days are there in 10 weeks

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There are 70 days in 10 weeks
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Step-by-step explanation:

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3 years ago

5. Find the first, fourth, and tenth terms of the arithmetic

Lubov Fominskaja [6]

The first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

Explanation:

The given rule for the arithmetic sequence is $A(n)=-6+(n-1)(\frac{1}{5} )$

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute $n=1,4,10$ in the general rule for the arithmetic sequence.

To find the first term, substitute $n=1$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(1)=-6+(1-1)(\frac{1}{5} )$

$A(1)=-6+(0)(\frac{1}{5} )$

$A(1)=-6$

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute $n=4$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(2)=-6+(4-1)(\frac{1}{5} )$

$A(2)=-6+(3)(\frac{1}{5} )$

$A(2)=\frac{-30+3}{5}$

$A(2)=\frac{-27}{5}$

Thus, the fourth term of the arithmetic sequence is $-\frac{27}{5}$

To find the tenth term, substitute $n=10$ in $A(n)=-6+(n-1)(\frac{1}{5} )$ , we get,

$A(10)=-6+(10-1)(\frac{1}{5} )$

$A(10)=-6+(9)(\frac{1}{5} )$

$A(10)=-6+\frac{9}{5}$

$A(10)=-\frac{21}{5}$

Thus, the tenth term of the arithmetic sequence is $-\frac{21}{5}$

Hence, the first, fourth and tenth terms of the arithmetic sequence is $-6, -\frac{27}{5}$ and $-\frac{21}{5}$

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4 years ago