First you have to make them to a common denominator.
5/6 7/12 7/10
50/60 35/60 42/60
from these the greatest value is 35/60 because 35 parts of 60 is bigger than the rest.
so the answer is 7/12

8 0

4 years ago

I need help on numbers 11-16

agasfer [191]

Recall your SOH CAH TOA, or $\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}$

now, bear in mind that, the opposite side, to an angle, is the side right "in front of it", that is, if you were to put your eye on that angle, "the wall" you'd see on the other end, is the opposite side

the adjacent side, adjacent = next to, is the side that's touching the angle itself

and the hypotenuse, is always the slanted and longest side of all three

for example on 16, tangent of Z

if you put one eye on Z, you'll see on the other end, the side of 30 units
the adjacent side is the one touching Z, or the 40 units side

$\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(Z)=\cfrac{30}{40}
\\\\\\
\textit{which can be simplified to }\cfrac{3}{4}
$

and you'd do all others, the same way, using those ratios

use $\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad 
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}$

8 0

3 years ago

Solve the equation by using the basic properties of logarithms.<br> ln(x+4)=-1

amid [387]

Write both sides as powers of $e$ :

$\ln(x+4)=-1\implies e^{\ln(x+4)}=e^{-1}\implies x+4=e^{-1}$

since $b^{\log_ba}=a$ . Then

$x=e^{-1}-4\approx-3.63$

5 0

3 years ago

Read 2 more answers

Pedro was curious if circles a and b with x and y Being on a circle respectively with similar so he tried to math one figure ont

tatuchka [14]

Answer:

The strategy Pedro tend to do isn't worth, because using rigid transformation to place on figure onto the other won't prove that those figures are similar, but it would prove if they are congruent.

Remember that similarity is about proportionality between dimensions of those figures, so it's not about congruence. That means the figures won't match when we place one onto the other.

To prove similarity, Pedro should divide each pair of corresponding coordinate, if all divisions give the same result, that means those figures are similar.

<h3>Therefore, Pedro is not right.</h3>

3 0

3 years ago