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Alja [10]
4 years ago
15

Elena said the equation 9x+15=3x+15 has no solutions because 9x is greater than 3x. Do you agree with Elena? Explain your reason

ing.
Mathematics
2 answers:
baherus [9]4 years ago
8 0

Answer:

Disagree

Step-by-step explanation:

You can solve this problem by doing as follows:

9x+15=3x+15

-3x        -3x

=

6x+15=15

-15        -15

=

6x=0

Although x=0, x=0 is still a solution to the problem.

erik [133]4 years ago
5 0

Answer:

yes

Step-by-step explanation:

any number times nine will be greater if you times the same number by three,

ex:

9(5)+15=60

3(5)+15=30

9x+15 > 3x+15

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4 5/6 - 4 1/3 in simplest form
babunello [35]

\huge\text{Hey there!}

\mathsf{4\dfrac{5}{6}-4\dfrac{1}{3}\ =}

\mathsf{4\dfrac{5}{6}=\bf \dfrac{29}{6}}

\mathsf{4\dfrac{1}{3}=\bf \dfrac{13}{3}}

\mathsf{\dfrac{29}{6}-\dfrac{13}{3}}

\mathsf{\dfrac{(29\times3)-(13\times6)}{6\times3}}

\mathsf{29\times3 = \bf 87}\leftarrow \large\textsf{\underline{NUMERATOR}}

\mathsf{13\times6= \bf 78}\leftarrow\large\textsf{\underline{NUMERATOR}}

\mathsf{\dfrac{\bf 87-78}{6\times3}}

\mathsf{87-78=\bf 9}\leftarrow\large\textsf{\underline{NUMERATOR}}

\mathsf{\dfrac{\bf 9}{6\times3}}

\mathsf{6\times3 = \bf 18}\leftarrow\large\textsf{\underline{DENOMINATOR}}

\mathsf{\dfrac{9}{\bf 18}}

\mathsf{\dfrac{9\div9}{18\div9}}

\mathsf{9\div9=\bf 1}\leftarrow \large\textsf{NUMERATOR}

\mathsf{\dfrac{\bf 1}{18\div9}}

\mathsf{18\div9= \bf 2}\leftarrow\large\textsf{DENOMINATOR}

\mathsf{ = \dfrac{1}{\bf 2}}

\boxed{\boxed{\large\text{Answer: }\mathsf{\bf \dfrac{1}{2}}}}\huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

6 0
3 years ago
The length of a rectangular garden is 10 feet longer than its width. if the garden's perimeter is 184 feet, what is the area of
Annette [7]
Width = w
length = w + 10
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2w + 2(w+10) = 184
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w = 164/4 = 41

w = 41
l = 51

Area = wl
Area = 41×51 = 2091 Sq ft
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svp [43]
Subtract like terms.
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