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2 years ago

11

Carlos Measured the link of a book to be 21 cm the actual length of a book is 23cm which of these is closest to the percent erro

r for Diego's measurements

1 answer:

Wewaii [24]2 years ago

5 0

Answer: 8.70%

Step-by-step explanation:

Percent error is calculated as:

= (Accepted value - Experimental value) / Accepted value × 100

= (23 - 21) / 23 × 100

= 2/23 × 100

= 8.70%

The percentage error is 8.70%

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Parallelogram PARL is similar to parallelogram WXYZ. If AP = 15, PL = 40, and WZ = 120, find the value of c .50 110 6 45

vova2212 [387]

Answer:

Option D. c = 45 will be the answer.

Step-by-step explanation:

In the figure attached there are two similar parallelograms PARL and WXYZ.

In which AP = 15, PL = 40, WZ = 120 and WX = c

By the property of similarity, corresponding sides of the given parallelograms will be in the same ratio.

$\frac{PL}{WZ}=\frac{PA}{WX}$

$\frac{40}{120}=\frac{15}{c}$

c = $\frac{15\times 120}{40}$ [By cross multiplication]

c = 45

Therefore, Option D. c = 45 will be the answer.

4 0

2 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

2 years ago

What do you know abou the graph of y=5 ?

Alisiya [41]

Answer:

Step-by-step explanation:

Solution : The graph of y=5 is a line parallel to the x-axis at a distance of 5 units from the origin

hope that help

mark me brillllll

3 0

2 years ago

What is the inverse of y=5x-5

Leya [2.2K]

Answer:the answer is in the picture below

Step-by-step explanation:plz give brainlist

8 0

1 year ago

What is the greatest common factor of 18 and 3?

-Dominant- [34]

3 is your greatest

3/3 = 1

18/3 = 6

hope this helps

8 0

2 years ago