All categories

3 years ago

What is the length of the diagonal of a 10 cm by 15 cm rectangle?

Mathematics

1 answer:

wariber [46]3 years ago

6 0

Answer:

18.03 cm

Step-by-step explanation:

The two sides of the rectangle are the legs of a right triangle and the diagonal is the hypotenuse. We can use the Pythagorean theorem

a^2 + b^2 = c^2

10^2 +15^2 = c^2

100 +225 = c^2

325 = c^2

Take the square root of each side

sqrt(325) = sqrt(c^2)

18.02775638 =c

Rounding to the nearest hundredth

18.03 =c

You might be interested in

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

6 0

3 years ago

please help me with this math problem, the question is on the picture that i attached, ill mark brainliest if the answer is corr

dsp73

Answer:

cos z = 35 / 37

Step-by-step explanation:

cos z is the ratio of the adjacent side of a triangle (the side adjacent to angle z to the hypotenuse:

cos z = (adj. side) / (hypotenuse)

Here we have cos z = 35 / 37

7 0

3 years ago

Answer for brainilest 11 points

TEA [102]

Answer:

729 is 3 to the power of 6

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

jillians school is selling tickects for a play. the tickets cost $10.50 for adults and $3.75 for students

Natali5045456 [20]

Answer:

What is the question?

Step-by-step explanation:

We need to know what problem to solve with the information provided.

4 0

2 years ago

Customers of a phone company can choose between two service plans for long distance calls. The first plan has a 26

Wittaler [7]

26+0.08x=8+0.13x
26-8=0.13x-0.08x
18=0.05x
x=18/0.05=360 min

6 0

2 years ago

Read 2 more answers