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sweet [91]
3 years ago
10

PLEASE HELP EM THANKS Solve |c - 2| = 10.

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
4 0
|c-2|=10\\
c-2=10 \vee c-2=-10\\
c=12 \vee c=-8
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Find the zero of the following function: f(x)=2/3(x)+6
elixir [45]
X=-9.
explanation: You want to set the equation equal to 0, so 0=2/3x+6. Subtract 6 from each side... -6=2/3x. then multiply each side by the reciprocal of 2/3, or 3/2. The equation is now 6(3/2) = 2/3x(3/2). Knowing that the reciprocal of any number times itself =1, we can conclude that -18/2, or -9=1x.
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3 years ago
can someone please help I don’t get it and I just want answers I have been trying to solve this for 1 hour now
dusya [7]

Answer:

1. y + 10 - 3/2y = -y/2 + 10

2. 2r+ 7r-r - 9 = 8r - 9

3.  7 + 4p-5+p+2q = 2 + 5p + 2q

Step-by-step explanation:

basically you can add terms that have the same variable

integers can be added together, Xs can be added, Zs, Ys, As, Bs, Cs, you get the point

1. y + 10 - 3/2y = -y/2 + 10

2. 2r+ 7r-r - 9 = 8r - 9

3.  7 + 4p-5+p+2q = 2 + 5p + 2q (do not add different variables p and q ) together

try 4-6 on your own to get this skill down, if you need help with those just let me know

7 0
3 years ago
I need help finding the area
aniked [119]
<h3>Given :</h3>
  • Base of triangle = 7 yd
  • Height of triangle = 10 yd

\\  \\

<h3>To find:</h3>
  • Area of triangle

\\  \\

We know:-

When base and height of triangle is given we use this formula:

\bigstar \boxed{ \rm Area \: of \: triangle =  \frac{base \times height}{2} }

\\  \\

So:-

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{base \times height}{2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 10}{2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times2 }{2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times\cancel2 }{\cancel2}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =  \dfrac{7 \times 5 \times1 }{1}  \\

\\  \\

\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5

\\  \\

\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2}  \\

\\  \\

\therefore  \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} {  \red{\bf{yd} }^{ \red2} }}

\\  \\

<h3>know more :-</h3>

\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}]

7 0
2 years ago
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Anarel [89]

Answer:

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Julli [10]

Answer:

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