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2 years ago

8

The value of 1 in 375,081 is 10

2 answers:

Setler [38]2 years ago

7 0

The value of the 1 is in the ones place so its 1

Zolol [24]2 years ago

6 0

No....the value of 1 in 375,081 is 1....because it is in the ones place

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I need help with number 8

Mumz [18]

Answer:

If the underlined digit is 5 then it is Fifty thousand (50,000)

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2 years ago

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Which equation best describes the line of best fit?

Andrej [43]

The y-value changes from 10 to 70 as the x-value changes from 0 to 10. That makes the slope (70 -10)/(10 -0) = 6, and the y-intercept = 10.

Selection A is the appropriate equation.

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3 years ago

Suppose r(t)=costi+sintj+3tk represents the position of a particle on a helix, where z is the height of the particle above the g

Ilia_Sergeevich [38]

a. The $\vec k$ component tells you the particle's height:

$3t=16\implies t=\dfrac{16}3$

b. The particle's velocity is obtained by differentiating its position function:

$\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k$

so that its velocity at time $t=\frac{16}3$ is

$\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k$

c. The tangent to $\vec r(t)$ at $t=\frac{16}3$ is

$\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t$

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2 years ago

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

Thus the maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

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3 years ago

4 (b-6)+19 simplify the expression

BlackZzzverrR [31]

4b - 6b + 19
Answer : -2b + 19

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3 years ago