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3 years ago

Two consecutive integers add to 51. what are the integers?

Mathematics

2 answers:

Nataly [62]3 years ago

7 0

25 and 26...........

Vedmedyk [2.9K]3 years ago

4 0

Answer: 25 and 26

Step-by-step explanation: This problem states that the sum of two consecutive integers is 51 and it asks us to find the integers. Two consecutive integers can be represented as followed.

X ⇒ <em>first integer</em>

X + 1 ⇒ <em>second integer </em>

Since the sum of our two con consecutive integers is 51, we can set up an equation to represent this.

X + X + 1 = 51

Now, we can simplify on the left side of our equation.

2x + 1 = 51

-1 -1 ← subtract 1 on both sides

2x = 50

÷2 ÷2 ← divide both sides by 2

X = 25

X + 1 = 26

Therefore, our two consecutive integers are 25 and 26.

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Step-by-step explanation:

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3 years ago

Read 2 more answers

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

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or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

3 years ago

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natima [27]

Answer:

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Step-by-step explanation:

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2 years ago