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3 years ago

A digital scale reports a 10 kg weight as weighing 9.998 kg. Which of the following is true?

2 answers:

vladimir2022 [97]3 years ago

6 0

A)

Accurate = close to the 'true' value, and 9.998 is close to 10. Imagine a scale that has the zero point wrong as says 8.998, it would not be accurate, but still precise (several digits known)

Precise = several digits known. I would say to know it until grams is quite a good precision (the 9.998 has 4 digits)

Semmy [17]3 years ago

6 0

Answer:

it's C . I just did this myself : P

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Adding and subtracting what is the solution to x = 4 = 12

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The answer would be 3 because 3✖️4= 12

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4 years ago

What are the weights (in Newton’s) for the given masses 22.1,33.5,41.3,59.2,78

Blababa [14]

Answer:

1. 216.8 N

2.328.635 N

3. 405.153 N

4. 580.752 N

5. 765.18 N

Step-by-step explanation:

1. Let the mass is 22.1 kg, then the weight in Newton will be (22.1 × 9.81) = 216.8 N {Where acceleration due to gravity is 9.81 m/sec²}

2. Let the mass is 33.5 kg, then the weight in Newton will be (33.5 × 9.81) = 328.635 N {Where acceleration due to gravity is 9.81 m/sec²}

3. Let the mass is 41.3 kg, then the weight in Newton will be (41.3 × 9.81) = 405.153 N {Where acceleration due to gravity is 9.81 m/sec²}

4. Let the mass is 59.2 kg, then the weight in Newton will be (59.2 × 9.81) = 580.752 N {Where acceleration due to gravity is 9.81 m/sec²}

5. Let the mass is 78 kg, then the weight in Newton will be (78 × 9.81) = 765.18 N {Where acceleration due to gravity is 9.81 m/sec²}

(Answer)

4 0

3 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

3 years ago

Someone please help me with two answer choices

otez555 [7]

Answer:

It is the first and the last three.

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3 years ago