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3 years ago

Question Progress Homework Progress 4/ Work out the size of angle x. 105 140

Mathematics

1 answer:

agasfer [191]3 years ago

5 0

Answer:

Step-by-step explanation:

Sum of angles of a triangle = 180

x + (180 - 105) + (180 -140) = 180

x + 75 + 40 = 180

x = 180 -75 - 40

x = 65

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Roman55 [17]

Answer:

0.25

Step-by-step explanation:

the unit rate in this case is the cost per line, you decide 2.00 by 8

2.00/8=0.25

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2 years ago

208 divi by 6.4 What is the quotient for the equation above?

andrey2020 [161]

Answer:

Step-by-step explanation:

208 divided by 6.4 = 32.5

8 0

3 years ago

Read 2 more answers

How to solve for<br> LN and what are the variables

Lesechka [4]

Answer:

v See below. v

Step-by-step explanation:

LM = MN

11x - 21 = 8x + 15

$3x-21=15\\3x=36\\$

x = 12

LM = 11(12) - 21 = 132 - 21 = 111

MN = 8(12) + 15 = 96 + 15 = 111

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3 years ago

The measurenicnt of the circumference of a circle is found to be 56 centimeters. The possible error in measuring the circumferen

BartSMP [9]

Answer:

(a) Approximate the percent error in computing the area of the circle: 4.5%

(b) Estimate the maximum allowable percent error in measuring the circumference if the error in computing the area cannot exceed 3%: 0.6 cm

Step-by-step explanation:

(a)

First we need to calculate the radius from the circumference:

$c=2\pi r\\r=\frac{c}{2\pi } \\c=8.9 cm$

I leave only one decimal as we need to keep significative figures

Now we proceed to calculate the error for the radius:

$\Delta r=\frac{dt}{dc} \Delta c\\\\\frac{dt}{dc} = \frac{1}{2 \pi } \\\\\Delta r=\frac{1}{2 \pi } (1.2)\\\\\Delta r= 0.2 cm$

$r = 8.9 \pm 0.2 cm$

Again only one decimal because the significative figures

Now that we have the radius, we can calculate the area and the error:

$A=\pi r^{2}\\A=249 cm^{2}$

Then we calculate the error:

$\Delta A= (\frac{dA}{dr} ) \Delta r\\\\\Delta A= 2\pi r \Delta r\\\\\Delta A= 11.2 cm^{2}$

$A=249 \pm 11.2 cm^{2}$

Now we proceed to calculate the percent error:

$\%e =\frac{\Delta A}{A} *100\\\\\%e =\frac{11.2}{249} *100\\\\\%e =4.5\%$

(b)

With the previous values and equations, now we set our error in 3%, so we just go back changing the values:

$\%e =\frac{\Delta A}{A} *100\\\\3\%=\frac{\Delta A}{249} *100\\\\\Delta A =7.5 cm^{2}$

Now we calculate the error for the radius:

$\Delta r= \frac{\Delta A}{2 \pi r}\\\\\Delta r= \frac{7.5}{2 \pi 8.9}\\\\\Delta r= 0.1 cm$

Now we proceed with the error for the circumference:

$\Delta c= \frac{\Delta r}{\frac{1}{2\pi }} = 2\pi \Delta r\\\\\Delta c= 2\pi 0.1\\\\\Delta c= 0.6 cm$

5 0

3 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$

we know that $cos^2A+sin^2A=1$ , so we have that

$f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)$

$lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19$

So f(0,0)=ln19.

8 0

3 years ago