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tester [92]
3 years ago
6

What divided by 5 equals 7

Mathematics
2 answers:
o-na [289]3 years ago
3 0
35 because multiplying 5 and 7 will equal 35, therefore 35/5=7.
Sergio039 [100]3 years ago
3 0

Answer:

35

Step-by-step explanation:

35 is a multiple of both 5 and 7. So if you multiply both of those numbers together, you get 35. Divide 35 by 5 and you get 7.

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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}]

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Find all the critical points of the function <img src="https://tex.z-dn.net/?f=%20f%28x%29%3D%20%28x%2B1%29%2F%28x-3%29%20" id="
yulyashka [42]

The function

... y = 1/x

has derivative

... y' = -1/x²

which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.

Your function

... y = (x+1)/(x-3)

can be written as

... y = 1 +4/(x-3)

which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).

The critical point is x=3.

The function is decreasing on (-∞, 3) ∪ (3, ∞), increasing nowhere.

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