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BARSIC [14]
3 years ago
15

Which expression is equivalent to (3b +2r) + (4b + r )

Mathematics
2 answers:
VladimirAG [237]3 years ago
6 0

Answer:

Step-by-step explanation:

The one that looks most closely to this.

(3b + 2r) + (4b + r)                 Remove the brackets.

3b + 2r + 4b + r                     Rewrite so the terms are close to each other.

3b + 4b + 2r + r                     Combine

7b + 3r    or

3r + 7b

One of the two of these can be given as an answer. Please give choices if you want me to give an exact answer.

givi [52]3 years ago
5 0

\implies {\blue {\boxed {\boxed {\purple {\sf { \: 7b + 3r}}}}}}

\sf \bf {\boxed {\mathbb {Step-by-step\:explanation:}}}

\: (3b + 2r) + (4b + r)

➼\: 3b + 2r + 4b + r

Combining like terms, we have

➼\: 3b + 4b + 2r + r

➼\: 7b + 3r

\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}

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Answer:

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Step-by-step explanation:

we have

0=9(x^{2} +6x)-18

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18=9(x^{2} +6x)

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11=(x^{2}+6x+9)

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The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter
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Answer:

(1\frac{1}{2},1) - False

(4,6) - True

(18,12) -- False

(0,0) -- True

(2\frac{1}{2},3\frac{3}{4}) -- True

Step-by-step explanation:

The points are

(1\frac{1}{2},1) , (4,6), (18,12), (0,0) and (2\frac{1}{2},3\frac{3}{4}) ---- missing from the question

Given

y = 1\frac{1}{2}x

Required

Determine if each of the points would be on y = 1\frac{1}{2}x

To do this, we simply substitute the value of x and of each point in y = 1\frac{1}{2}x.

(a) (1\frac{1}{2},1)

In this case;

x = 1\frac{1}{2} and y = 1

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 1\frac{1}{2}

y = \frac{3}{2} * \frac{3}{2}

y = \frac{9}{4}

y = 2\frac{1}{4}

<em>The point </em>(1\frac{1}{2},1)<em>  won't be on the graph because the corresponding value of y for </em>x = 1\frac{1}{2}<em> is </em>y = 2\frac{1}{4}<em></em>

(b) (4,6)

In this case;

x = 4

y = 6

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 4

y = \frac{3}{2} * 4

y = \frac{3* 4}{2}

y = \frac{12}{2}

y = 6

<em>The point </em>(4,6)<em>  would be on the graph because the corresponding value of y for </em>x = 4 is y = 6

(c) (18,12)

In this case:

x = 18;y = 12

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 18

y = \frac{3}{2} * 18

y = \frac{3* 18}{2}

y = \frac{54}{2}

y = 27

<em>The point </em>(18,12)<em>  wouldn't be on the graph because the corresponding value of y for </em>x = 18<em> is </em>y = 12<em></em>

(d) (0,0)

In this case;

x =0; y = 0

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 0

y = 0

<em>The point </em>(0,0)<em>  would be on the graph because the corresponding value of y for </em>x = 0 is y = 0

(e) (2\frac{1}{2},3\frac{3}{4})

In this case:

x = 2\frac{1}{2}; y = 3\frac{3}{4}

y = 1\frac{1}{2}x becomes

y = 1\frac{1}{2} * 2\frac{1}{2}

y = \frac{3}{2} * \frac{5}{2}

y = \frac{15}{4}

y = 3\frac{3}{4}

<em>The point </em>(2\frac{1}{2},3\frac{3}{4}) <em>  would be on the graph because the corresponding value of y for </em>x = 2\frac{1}{2} is y = 3\frac{3}{4}

3 0
3 years ago
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