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Ulleksa [173]
4 years ago
8

Forty-two divided by seven plus the quantity three divided by six 1. Write the numerical expression. 2. Evaluate within parenthe

ses. 3. There are no exponents to evaluate. 4. Multiply and divide from left to right. 5. Add and subtract from left to righ
Mathematics
2 answers:
lorasvet [3.4K]4 years ago
5 0

Answer:

1. (d)  42 div 7+(3 div 6)

2. (b)  42 div 7+ 0.5

4. (c) 6+0.5

5. (b) 6.5

Step-by-step explanation:

your welcome

Leona [35]4 years ago
4 0

Answer:

<h2>6.5</h2>

Step-by-step explanation:

Given Forty-two divided by seven plus the quantity three divided by six, the equivalent numerical expression will be;

\frac{42}{7} + \frac{3}{6}

To evaluate the numerical expression, we will find the LCM of the denominator

\frac{42}{7} + \frac{3}{6} = \frac{6(42)+7(3)}{42}\\ = \frac{252+21}{42}\\= \frac{273}{42}\\= 6.5

The value of the expression 6.5

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Answer:

1. 6.75x or (6.75, 1)

2. He buys 9 sandwiches.

Step-by-step explanation:

For the first one, we know that he pays $6.75 for each sandwich, so if x equals the number of sandwiches, when written out it would be 6.75x

For the second one, we have to divide it up and solve one step at a time, and when we start solving them, we will find the pieces to figure out the other problems. Let's start with how much he paid. We know that he paid $80, but we have to remember that he got change back. So to find out how much he truly paid, we subtract the 19.25 from the 80. When we do this, we get 60.75. Now, to find out how many sandwiches he got, we can use how much he paid for each one. We know that he paid $6.75 for each one, and in total he spent $60.75. So to find out, we can simply divide 6.75 into 60.75, and we get 9. So now we know that he bought 9 sandwiches.

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3 years ago
Which expression is equivalent to StartFraction 2 a + 1 Over 10 a minus 5 Endfraction divided by StartFraction 10 a Over 4 a squ
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For this case we must simplify the following expression:

\frac {\frac {2a + 1} {10a-5}} {\frac {10a} {4a ^ 2-1}} =

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\frac {(4a ^ 2-1) (2a + 1)} {10a (10a-5)} =

We apply distributive property in the numerator and we take common factor 5 in the denominator:

\frac {8a ^ 3 + 4a ^ 2-2a-1} {10a (5 (2a-1))} =

We factor the numerator:

\frac {(2a + 1) ^ 2 (2a-1)} {10a (5 (2a-1))} =\\\frac {(2a + 1) ^ 2 (2a-1)} {50a (2a-1)} =

We simplify:

\frac {(2a + 1) ^ 2} {50a}

Answer:

\frac {(2a + 1) ^ 2} {50a}

Option D

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3 years ago
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Find tan2θ if θ terminates in Quadrant IV and cosθ = 3/5.
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\cos( \alpha )  =  \frac{3}{5}   \\

{sin}^{2}( \alpha )  = 1 -  {cos}^{2}  ( \alpha )

{sin}^{2}( \alpha )  = 1 -  ({ \frac{3}{5} })^{2} \\

{sin}^{2}( \alpha )   = 1 -  \frac{9}{25}  \\

{sin}^{2}( \alpha )   =  \frac{25}{25}  -  \frac{9}{25}  \\

{sin}^{2}( \alpha ) =  \frac{16}{25}   \\

sin( \alpha )= ± \sqrt{ \frac{16}{25} }  \\

In Quadrant IV , sin is negative .

Thus ;

sin( \alpha ) =  -  \sqrt{ \frac{16}{25} }  \\

\sin( \alpha )  =  -  \frac{4}{5}  \\

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

\tan(2 \alpha )  =  \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) }  \\

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

\sin(2 \alpha ) = 2. \sin( \alpha ).  \cos( \alpha )

\sin(2 \alpha )  = 2 \times ( -  \frac{4}{5} ) \times ( \frac{3}{5} ) \\

\sin(2 \alpha )  =  -  \frac{24}{25}  \\

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\cos(2 \alpha ) =  {cos}^{2}( \alpha ) -  {sin}^{2}( \alpha )

\cos(2 \alpha ) =  ({ \frac{3}{5} })^{2} -  ({ -  \frac{4}{5} })^{2}   \\

\cos(2 \alpha )  =  \frac{9}{25}  -  \frac{16}{25}  \\

\cos(2 \alpha )  =  -  \frac{7}{25}  \\

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\tan(2 \alpha )  =  \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) }  \\

\tan(2 \alpha )  =  \frac{ -  \frac{24}{25} }{ -  \frac{7}{25} }  \\

\tan(2 \alpha ) =  -  \frac{24}{25}    \div   -  \frac{7}{25}  \\

\tan(2 \alpha )  =  -  \frac{24}{25}  \times   -  \frac{25}{7}  \\

\tan(2 \alpha )  =  \frac{24}{7}  \\

Thus the correct answer is (( C )) .

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