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4 years ago

Round to the nearest ten and hundred 829

Mathematics

2 answers:

Marysya12 [62]4 years ago

6 0

Nearest 10th = 830
Nearest 100th = 800

Musya8 [376]4 years ago

4 0

To the nearest ten, 829 would become 830 because the ones digit (9) is closer to 30 than it is to 20. So we would round up.

To round to the nearest hundred, we look at the tens and ones digit. 29 is less than 50 so we would have to round down to 800 instead of rounding up to 900.

Hope that helps :)

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andreev551 [17]

B) $x =4.28$

C) volume = $135.25cm^3$ .

Step-by-step explanation:

Here we have , The base of an open rectangular box is of length (2x+5) cm and width x cm. The area of this base is 58cm^2. The height of the open box is (x-2) cm. We need to find the following :

a) show that 2x^2+5x-58=0

We know that area of rectangle = $length(width)$

⇒ $Area = length(width)$

Putting values according to question we get :

⇒ $58 = (2x+5)(x)$

⇒ $58 = (2x^2+5x)$

⇒ $2x^2+5x - 58 = 0$

b) solve the equation in the question above, giving your answer to 2 decimal places

Solving this equation :

⇒ $2x^2+5x - 58 = 0$

By quadratic formula

⇒ $x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $x = \frac{-5 \pm \sqrt{5^2-4(2)(-58)} }{2(2)}$

⇒ $x = \frac{-5 \pm 22.11}{4}$

⇒ $x = \frac{-5 + 22.11}{4} , x = \frac{-5 - 22.11}{4}$ { Since side can't be negative ! }

⇒ $x =4.28$

c) calculate the volume of the box, stating units of you answer

Since x = 4.28 , other sides are

$2x+5=2(4.28)+5=13.56\\x-2=4.28-2=2.28$

We know that volume is given by :

⇒ $length(width)(height)$

⇒ $13.86(4.28)(2.28)$

⇒ $135.25cm^3$

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x can also be anything between 2 and -4, so B is also incorrect.

That leaves C our answer.

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