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2 years ago

Simplify the expression. (5+3)×4 8 17 23 32

Mathematics

1 answer:

Airida [17]2 years ago

5 0

Answer:

Step-by-step explanation:

BIDMAS

(5 + 3) = 8

8 x 4 = 32

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Domain is [-10,∞)
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I think, it should be like this.

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3 years ago

I need a answer to the picture provided by me.

Zepler [3.9K]

Answer:

the first one is -(25/2) OR -12.5

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2 years ago

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Raymond just got done jumping at Super Bounce Trampoline Center. The total cost of his session was $43.25 He had to pay a $7 ent

Travka [436]

Answer:

the number of the minutes 29

Step-by-step explanation:

because the if we take the cost from the session it will be 36.25 and divide by 1.25 because it takes every minute 1.25 so it will be 29

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2 years ago

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Rich measured the height of a desk to be 80.7 cm. The actual height of the desk is 80 cm. What is Rich's percent error in this c

Hitman42 [59]

Answer:

0.9%

Step-by-step explanation:

We have been given that Rich measured the height of a desk to be 80.7 cm. The actual height of the desk is 80 cm.

We will use percentage error formula to solve our given problem.

$\text{Percentage error}=\frac{\text{Experimental value-Actual value }}{\text{Actual actual}}\times 100$

$\text{Percentage error}=\frac{80.7-80}{80}\times 100$

$\text{Percentage error}=\frac{0.7}{80}\times 100$

$\text{Percentage error}=0.00875\times 100$

$\text{Percentage error}=0.875\approx 0.9$

Therefore, Rich's percent error in calculation is 0.9%.

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3 years ago

A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is

Monica [59]

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

$l^2 = h^2 +d^2$

$l^2 = 6^2 + d^2$

$l^2 = 36 + d^2$

We relate to: $2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}$

From the question;

l = 34,

So to find $\dfrac{dd}{dt}$ , we get;

$d = \sqrt{l^2 - 36}$

$d = \sqrt{34^2 - 36}$

$d = \sqrt{1156- 36}$

$d = \sqrt{1120}$

d = 33.46

So, we have:

$\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}$

$\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5$

$\dfrac{dd}{dt}= 1.016 \times - 5$

$\dfrac{dd}{dt}=-5.08 \ ft/sec$

4 0

2 years ago