Help! Will mark brainliest answer!!

Mathematics

1 answer:

Gnom [1K]3 years ago

8 0

PART A:

The option that would be equal to figure ABC would be A scalene triangle with only one angle that measures 80* because the other angle is not shown in this phrase so we could say that the other angle is 50* then it would be equal to figure ABC, and a scalene triangle is a triangle with three uneven, unequal sides,

Therefore the answer would be,

C. A scalene triangle with only one angle that measures 80*

You might be interested in

A boy's mother was 32 years old when he was born. Let a represent the boy's age and m represent the mother's age.

LenKa [72]

see the picture attached

5 0

3 years ago

Find the nth term of this number sequence<br> 7, 10, 13, 16, ...<br><br> help please !!

LenKa [72]

Answer:

$a_{n}$ = 3n + 4

Step-by-step explanation:

there is a common difference between consecutive terms , that is

10 - 7 = 13 - 10 = 16 - 13 = 3

this indicates the sequence is arithmetic with nth term

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

here a₁ = 7 and d = 3 , then

$a_{n}$ = 7 + 3(n - 1) = 7 + 3n - 3 = 3n + 4

8 0

2 years ago

Read 2 more answers

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$