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Maurinko [17]
3 years ago
7

Round to nearest centimeter 80.5 cm is how many cm?

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0
Its 81 centimeters because you round 5 and over. So nearest centimeter means 2 whole numbers which is 81
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WILL GIVE BRAINLIEST TO CORRECT ANSWER
MA_775_DIABLO [31]

Answer: 80

Step-by-step explanation:

Multiple 8 x 10

7 0
3 years ago
Why would understanding perpendicular and parallel lines be extremely important to someone in construction? Provide an example o
Tems11 [23]

Understanding perpendicular and parallel lines are extremely important, especially in the engineering field when building infrastructures or houses because they need to calculate how they will keep a building stable. For example, bridges we see have a lot of perpendicular and parallel lines, that's because without those the bridge can't hold itself up, it needs support from the metals that are parallel and perpendicular.

hope this helps!!

5 0
3 years ago
Divide 124 in the ratio of 3:1 plsssss
UNO [17]
The answer is 93 to 31 :)
6 0
4 years ago
Read 2 more answers
1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help
Gnesinka [82]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify :- 1 + - w² + 9w.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf1 + - w ^ { 2 } + 9 w

Quadratic polynomial can be factored using the transformation \sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \sf x_{1} and x_{2} are the solutions of the quadratic equation \sf \: ax^{2}+bx+c=0.

\large \sf-w^{2}+9w+1=0

All equations of the form \sf\:ax^{2}+bx+c=0 can be solved using the quadratic formula: \sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)}  \\

Square 9.

\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)}  \\

Multiply -4 times -1.

\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)}  \\

Add 81 to 4.

\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)}  \\

Multiply 2 times -1.

\large \sf \: w=\frac{-9±\sqrt{85}}{-2}  \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is plus. Add -9 to \sf\sqrt{85}.

\large \sf \: w=\frac{\sqrt{85}-9}{-2}  \\

Divide -9+ \sf\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is minus. Subtract \sf\sqrt{85} from -9.

\large \sf \: w=\frac{-\sqrt{85}-9}{-2}  \\

Divide \sf-9-\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}}  \\

Factor the original expression using \sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \sf\frac{9-\sqrt{85}}{2}for \sf\:x_{1} and \sf\frac{9+\sqrt{85}}{2} for \sf\:x_{2}.

\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

\large \sf \: 1 + -  w {}^{2}  + 9w \\  =\large  \boxed{\bf \: 1 -  {w}^{2}   + 9w}

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0
3 years ago
3) 14, 9, 20, 5, 17, 13 mean: <br>median: <br>mode: <br>range:​
Korolek [52]

Answer:

mean: 13

range: 15

there is no mode

median: 13.5

Step-by-step explanation:

mean= 14+9+20+5+17+13 = 78

78÷6= 13

range = highest number minus the smallest number

20-5 = 15

the mode is the number that occurs the most

median= 13+14= 27

27÷2 = 13.5

7 0
3 years ago
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