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3 years ago

A triangle with three acute angles and no congruent sides

Mathematics

1 answer:

Sauron [17]3 years ago

8 0

The answer should be an acute scalene triangle.

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Triangle J is shown below. James drew a scaled version of Triangle J using a scale factor of 4 and labeled it

Novay_Z [31]

Answer:

The area of triangle K is 16 times greater than the area of triangle J

Step-by-step explanation:

we know that

If Triangle K is a scaled version of Triangle J

then

Triangle K and Triangle J are similar

If two triangles are similar, then the ratio of its areas is equal to the scale factor squared

Let

z -----> the scale factor

Ak ------> the area of triangle K

Aj -----> the area of triangle J

$z^{2}=\frac{Ak}{Aj}$

we have

$z=4$

substitute

$4^{2}=\frac{Ak}{Aj}$

$16=\frac{Ak}{Aj}$

$Ak=16Aj$

therefore

The area of triangle K is 16 times greater than the area of triangle J

4 0

3 years ago

Read 2 more answers

The base area of a cylinder is 200.96m2

Vitek1552 [10]

Answer:

Step-by-step explanation:

Volume of cylinder = base area * height

height of the cylinder = Volume of the cylinder ÷ base area

= 1205.76 ÷ 200.96

= 6 in

7 0

2 years ago

Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i

liberstina [14]

Answer:

ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

$\diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}$