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AveGali [126]
4 years ago
14

Which best describes the relationship between the two triangles below?

Mathematics
1 answer:
Nostrana [21]4 years ago
3 0

Answer:

They are SIMILAR triangles

Step-by-step explanation:

Given triangles MNL and FHG with two of their angles to be 51° and 36° respectively. Their third angle is expressed as:

180°-(51°+36°)

= 180°-87°

= 93°

Their third angle is 93°

Since both triangles have the same angles in them, both triangles are considered as similar triangle even though the value of their sides are different due to angle differences in both.

Note that individual triangles are scalene triangles since their angles are not the same but the both triangles are similar triangles based on their relationship.

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Please help me i really need help please
Contact [7]

Answer:  66.

explanation: Ok so right triangle: 1 right angle or 90° and 2 acute angles (Less than 90°). We know that one of the angles are equal to 24° and the right angle is equal to 90° degrees. Now what? We add those 2 values together. 90 + 24 = 114. In the background, a triangle is equal to 180°. So now that we have our 2 core values together, we subtract. The reason why is because this is a missing value type equation. In these types, (usually indicated by the word find) we can subtract 2 values. Our 2 values is 114 and 180. When we subtract both, we get 66.

7 0
3 years ago
What does 12 1/3 - 11 3/4
myrzilka [38]
2/3 is left over 

I hope this is the answer you are looking for.
6 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}}

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Crazy boy [7]

Answer:

Step-by-step explanation:

Be more specific, X can be a variable, or they can be coordinates, visa versa.

7 0
3 years ago
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