Confused on these 4 questions

98 POINTS MATH!

patriot [66]

Let $a$ be the length of the leg with one tick mark and $b$ the length of the leg with two tick marks.

In the upper triangle, the law of cosines says

$6^2=a^2+b^2-2ab\cos43^\circ$

In the lower triangle, it says

$4^2=a^2+b^2-2ab\cos(4y-5)^\circ$

Subtract the second equation from the first to eliminate $a^2+b^2$ :

$6^2-4^2=-2ab(\cos43^\circ-\cos(4y-5)^\circ)$

$10=ab(\cos(4y-5)^\circ-\cos43^\circ)$

$a$ and $b$ are lengths so they must both be positive. 10 is also positive, so in order to preserve the sign on both sides of this equation, we must have

$\cos(4y-5)^\circ-\cos43^\circ>0$

$\cos(4y-5)^\circ>\cos43^\circ$

Now we have to be a bit careful. If $x$ is an acute angle, then as $x$ gets larger, the value of $\cos x$ gets smaller. So if we have two angles $\theta$ and $\varphi$ , with $0^\circ$ , then we would have $\cos\theta>\cos\varphi$ .

This means in our inequality, taking the inverse cosine of both sides would reverse the inequality:

$(4y-5)^\circ$

We know that $(4y-5)^\circ$ is an angle in a triangle, so it must be some positive measure:

$(4y-5)^\circ>0^\circ\implies y>\dfrac54^\circ$

So we must have

$\dfrac54^\circ$

6 0

3 years ago

Segments CD, AE, and BF are medians of triangle ABC. What is the value of x?

Slav-nsk [51]

Step-by-step explanation:

The centroid divides the length of each median in 2:1 ratio.

The length of the part between the vertex and the centroid is twice the length between the centroid and the mid-point of the opposite side.

When CD is the median, C is the vertex and D is the midpoint of AE.

CG = 2 × DG

6.5= 2(3x- 11)

6.5 = 6x -22

6x= 6.5+22

6x= 28.5

x= 28.5/6

x= 4.75

optionB

7 0

3 years ago

1.6y+y-4/15y+1 1/6y=2 1/3

Advocard [28]

1.6y+y-4/15y+7/6y=21/3

We move all terms to the left:

1.6y+y-4/15y+7/6y-(21/3)=0

Domain of the equation: 15y!=0

y!=0/15

y!=0

y∈R

Domain of the equation: 6y!=0

y!=0/6

y!=0

y∈R