All categories

4 years ago

△ABC∼△LMN , AB is 8 inches, BC is 14 inches, and LM is 12 inches. What is MN ? Enter your answer in the box.

Mathematics

2 answers:

Mariana [72]4 years ago

8 0

Answer:

The measure of MN is 21 inches.

Step-by-step explanation:

It is given that △ABC∼△LMN.

The corresponding sides of two similar triangles are proportional.

$\frac{AB}{LM}=\frac{BC}{MN}$

$\frac{8}{12}=\frac{14}{MN}$

$8\times MN=14\times 12$

Divide both sides by 8.

$MN=\frac{168}{8}$

$MN=21$

Therefore we can say that if △ABC∼△LMN, then the measure of MN is 21 inches.

Ostrovityanka [42]4 years ago

5 0

AB/LM = BC/MN

8/12 = 14/MN

8(MN) = 12 * 14

8(MN) = 168

MN = 168/8

MN = 21 Hope this helps and please mark as brainliest!

You might be interested in

Round to the nearest 10 and hundred 663

vladimir2022 [97]

660 - nearest ten
700 - nearest hundred

4 0

3 years ago

Read 2 more answers

10. In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes

RSB [31]

Answer:

The Box 1 and Box 2 are the same mass.

Explanation:

It caused both to change speed because they have both the same mass.

7 0

3 years ago

What are indications of a computer hardware problem? Check all that apply.

alexira [117]

Step-by-step explanation:

the answer is the keyboard keys getting stuck, it's something you can touch and feel, thus it's an hardware problem

6 0

3 years ago

Read 2 more answers

Find the inverse of this 3x3 matrix. [1 5 2]

MissTica

Lets say the 3x3 Matrix is

M = [1 5 2 ]
[1 1 7 ]
[0 -3 7 ]

We apply the Gauss-Jordan elimination method (Procedure and result shown in the image below)

5 0

3 years ago

Remember to show work and explain. Use the math font.

MrMuchimi

Answer:

$\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}$

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================$

$1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}$

$\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)$

$--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)$

$--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}$

6 0

3 years ago

△ABC∼△LMN , AB is 8 inches, ​ BC ​ is 14 inches, and LM is 12 inches. What is ​​ MN ? Enter your answer in the box.

△ABC∼△LMN , AB is 8 inches, BC is 14 inches, and LM is 12 inches. What is MN ? Enter your answer in the box.