Given ΔGHJ, complete the following statements. The included angle of sides JG and GH is angle . Angle H is the opposite angle to side . The side opposite to ∠J is side .

Answers

Answer 1
Answer:

Answer with explanation:

First Draw ΔG H J, and name the vertices of triangle in same order from G,to H to J.

→→The included angle of sides JG and G H is angle G.

→→Angle H is the opposite angle to side G J.

→→ The side opposite to ∠J is side G H.

If you will interchange the vertices of triangle and write it in any order you will get the same result.There are 6 ways ,

ΔG H J, ΔH G J, ΔH JG, ΔG J H ,ΔJ HG, ΔJ G H.

Answer 2
Answer: G , JG, GH is the answer i think 

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Solve for a: a2-8a+12 which is the correct a.-2,-6 b.2,6. C.-3,-4 or d.3,4 for me is Ccc help pleas

Answers

a^2-8a+12 =0 \n \n(a-2)(a-6)=0 \n \na-2=0 \ \ or \ \ a-6 =0 \n \na=2 \ \ or \ \ a=6 \n \nAnswer : \ b. \ \ 2,6


Which is 23,578,000 written in scientific
notation?

Answers

Answer:

23,578,000 = 2.3578 *  {10}^(7)  \n

Find the value of x.  Round to nearest hundredth if necessary.

Answers

The side on the front of 30° is 1/2 of hypytenuse
Then 1/2 of 4=2 and the ather side with theorem of pytagore is squares root of(4^2-2^2)=sq .root of 12
This side(square root of 12) is 1/2 of hypotenuse then
Hypotenuse is 2*square root 12
X^2=(2*sq .root12)^2 -(square root 12)^2
X^2=48-12=36
X=6

The sum of the squares of two consecutive positive even integers is 340. what are the integers?

Answers

12 squared and 14 squared.
12 squared is 144 and 14 squared is 196
144 + 196 = 340

Check all the equations that apply to meet the Segment AdditionPostulate in segment OQ. OP=6X+5,PQ=4x and OQ=45

Answers

Answer:

Options (2) and (5)

Step-by-step explanation:

Given in the question,

Length of OQ = 45

OP = (6x + 5)

PQ = 4x

By segment addition postulate,

Length of a segment OQ = length of segment OP + length of segment PQ

45 = 6x + 5 + 4x

10x + 5 = 45

10x = 45 - 5

10x = 40

x = (40)/(10)

x = 4

Therefore, Option (2) and Option (5) are the correct options.

An airplane has begun its descent for a landing. When the airplane is 150 miles west of its destination, its altitude is 25,000 feet. When the airplane is 90 miles west of its destination, its altitude is 19,000 feet. If the airplane's descent is modeled by a linear function, where will the airplane be in relation to the runway when it hits ground level?

Answers

The airplane has descended (25,000 - 19,000) = 6,000 feet
while flying (150 - 90) = 60 miles.

If the descent is modeled by a linear function, then the slope
of the function is

             (-6000 ft) / (60 miles)  =  - 100 ft/mile .

Since it still has 19,000 ft left to descend, at the rate of 100 ft/mi,
it still needs to fly

           (19,000 ft) / (100 ft/mile) =  190 miles

to reach the ground.

It's located 90 miles west of the runway now.  So if it continues
on the same slope, it'll be 100 miles past the runway (east of it)
when it touches down.

I sure hope there's another airport there.

Answer:

airplane will over shoot the runway by 100 miles (for USA Test Prep people)

Step-by-step explanation:

slope =

25,000 - 19,000

150 - 90

= 100

y = mx + b

25,000 = 100(150) + b

25,000 - 15,000 = b

b = 10,000

thus,

y = mx + b

0 = 100x + 10,000

-10,000 = 100x

x = -100

When the plane is at ground level (y = 0), it will be 100 miles past the end of the runway.