Please help with these 2!! Thank you

SOMEONE PLEASE HELLPPPP

ipn [44]

Answer:

93.39

Step-by-step explanation:

So the sum of exterior angles of the convex octagon is: 360 degrees

This means if we add all the equations that represent each angle, we can set it equal to 360 and solve for x

$(x+14) + (2x-3) + (3x+8) + (3x+16) + (2x-17) + (3x-4) + (3x-12) + (6x)$

Group like terms

$(x+2x+3x+3x+2x+3x+3x+6x) + (14-3+8+16-17-4-12)$

Add like terms

$23x+2$

Now let's set the sum of exterior angles to 360

$23x+2 = 360$

Subtract 2 from both sides

$23x=358$

Divide both sides by 23

$x\approx 15.565$

So by looking at all these, it appears that 6x is the highest value, given that x is positive. The way I estimated, is approximately 15.5, whenever I saw an equation like x+14, I estimated it's about 2x, since 14 is not exactly, but close to 15.5. I did this with each polynomial given. You could also manually check each one

Original equation

6x

Subsitute

6(15.565)

Simplify

$93.39$

7 0

1 year ago

Bateman's jar of marbles contains 8 blue marbles. 6 yellow marbles, and 9 tiger eyes. Bateman pulls a blue marble without puttin

xeze [42]

Answer: no

Step-by-step explanation:

7 0

3 years ago

There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t

NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s text which one of the triangle's side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM. AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM (1)

1. Let AC=21 So AM=21/2=10.5 cm

So AB=10.5 cm as well. So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill. AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB. So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC 21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0

3 years ago

Write a situation that could be represented by the equation y=15x-3

Solnce55 [7]

A tree grows at a rate of 15 inches per year. However, three inches were chewed off by quasi-sadistic bunnies the first year it was planted. Fortunately, the bunnies were frightened off by the garden gnomes, and did not return. As a result the tree was able to grow unencumbered for the rest of its life.

8 0

3 years ago

Help me please Find the measure of angle QST

Oksi-84 [34.3K]

Answer:

Angle QST=52°+47°

=99°

3 0

2 years ago