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Paraphin [41]
3 years ago
11

Pls help due soon ..........

Mathematics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

8.5

Step-by-step explanation:

So substitute the vales for r and d

(4*2)

and |-6| means the absolute value of -6, which is just 6

so simplifying the equation would make it

(8+6.5) - 6

14.5-6

8.5 :)

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This morning I went to Larry's for breakfast.
Olin [163]

Answer:

9 dollars and 18 cents

Step-by-step explanation:

6 0
3 years ago
Whats the normal arm span for these heights? : 4'10,4'11,5'0,5'4,5'5,5,'7,5'8,5'9,5'10,5'11,6'0
Svetllana [295]

 

In adults, the arm span is approximately 5 cm greater than the height in adult males and 1.2 cm in adult females. To calculate the arm span for the heights given, we add 5cm to their height. The following are the results:

 

Height                   Arm Span Length (in cm)

4’10                        152.32

4’11                        154.86

5’0                          157.4

5’4                          167.56

5’5                          170.10

5’7                          175.18

5’8                          177.72

5’9                          180.26

5’10                        182.80

5’11                        185.34

6’0                          187.88

 

To add, the total measurement of the length from the furthermost part of an individual's arms to the other end when raised equidistant to the ground at shoulder height at a 90º angle is called the arm span or wingspan.

7 0
3 years ago
A city known for its temperature extremes started the day at -5 degrees Fahrenheit. The temperature increased by 78 degrees Fahr
charle [14.2K]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Starting temperature = - 5°F

Temperature at midday =increase in temperature by 78°F = (starting temperature + 78°F)

Temperature at nightfall = decrease in temperature by 32°F = ( temperature at midday - 32°F)

Temperature at midday:

(-5°F + 78°F) = 73°F

Temperature at nightfall:

(Temperature at midday - 32°F)

(73°F - 32°F) = 41°F

7 0
3 years ago
1
Reptile [31]

Answer:

It's the third option: (x^2 - 5)(x - 7).

Step-by-step explanation:

x^3 – 7x^2 – 5x + 35

= x^2(x - 7) - 5(x - 7)        The (x - 7) is common so we have:

(x^2 - 5)(x - 7).

6 0
3 years ago
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.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

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}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\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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