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3 years ago

12

A calculator displays a result as 1.3540980 107 kg. the estimated uncertainty in the result is Â±2%. how many digits should be i

ncluded as significant when the result is written down?

1 answer:

yan [13]3 years ago

3 0

<span>2 significant digits.

Let's see what the range of possible values you can have for 1.3540980 if your uncertainty is +/- 2% 2% of 1.3540980 = 0.02 * 1.3540980 = 0.027082 So the lowest possible value for your result is 1.3540980 - 0.027082 = 1.327016 The largest possible result is 1.3540980 + 0.027082 = 1.38117996 Notice that only the 1st 2 digits of the result match which is reasonable since a 2% error means that your result is only accurate to within 1 part in 50.</span>

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Write a polynomial function of least degree with the given zero. -1+ √ 2, √ 3

kirill115 [55]

Answer:

f(x) = $x^{4}$ + 2x³ - 4x² - 6x + 3

Step-by-step explanation:

Note that radical zeros occur in conjugate pairs, thus

- 1 + $\sqrt{2}$ is a zero then - 1 - $\sqrt{2}$ is also a zero

$\sqrt{3}$ is a zero then - $\sqrt{3}$ is also a zero

Thus the corresponding factors are

(x - (- 1 + $\sqrt{2}$ ) ), (x - (- 1 - $\sqrt{2}$ ) ), (x - $\sqrt{3}$ ), (x - (- $\sqrt{3}$ )), that is

(x + 1 - $\sqrt{2}$ ), (x + 1 + $\sqrt{2}$ ), (x - $\sqrt{3}$ ), (x + $\sqrt{3}$ )

The polynomial is then the product of the roots

f(x) = (x + 1 - $\sqrt{2}$ )(x + 1 + $\sqrt{2}$ )(x - $\sqrt{3}$ )(x + $\sqrt{3}$ )

= ((x + 1)² - ( $\sqrt{2}$ )²)((x² - ( $\sqrt{3}$ )²)

= (x² + 2x + 1 - 2)(x² - 3)

= (x² + 2x - 1)(x² - 3) ← distribute

= $x^{4}$ - 3x² + 2x³ - 6x - x² + 3

= $x^{4}$ + 2x³ - 4x² - 6x + 3

6 0

3 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago

Angles H and F are corresponding angles<br> True or False?

lutik1710 [3]

True! they are corresponding angles.

5 0

2 years ago

Read 2 more answers

Jenny was planning a trip to the United Arab Emirates. Before going, she did some research and learned that the exchange rate is

Blizzard [7]

4 dirhams is equal to 1 dollar so if you were to multiply each by 5 you would have 20 dirhams equaling 5 dollars

5 0

3 years ago

SOLVE USING ELIMINATION *20 POINTS*

V125BC [204]

1. No answer ( may be due to incorrect question)

2. x = -3

y = 4

3. x= 4

y = 2

4.x = -63/40

y= -19/20

Step 1

Multiply equation 1 by the coefficient of x in equation 2

Multiply equation 2 by the coefficient of x I'm equation 1

(after completing this step you will derive equation 3 and 4 )

Step 2

Subtract equation 4 from equation 3

Step 3

Divide both sides of the equation by the coefficient of y

Step 4

substitute your value for y in equation 1 or 2

(after this you will derive the values of x)

Note : This method is for the Elimination of x

I hope it helps

5 0

3 years ago