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Fiesta28 [93]
3 years ago
6

Need help please someone

Mathematics
1 answer:
salantis [7]3 years ago
4 0

The answer to 9 would be (-3,-4) and (2,6) because if you plug in -3 for x in either of the equations and plug in 4 for y in the equations, both sides of the equation are the same number. also, if you plug in 2 for x and 6 for y both sides of the equation equal each other. The answer to 10 is (-1,1) and (7,33). if you plug them in they are equal:)

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An aircraft has a total time in service of 468 hours. The Airworthiness Directive given was initially complied with at 454 hours
Georgia [21]

Answer:

440 hours

Step-by-step explanation:

In this case the aircraft has completed one cycle of service in the first 454 hours of service. Then making the difference between the hours of service and the hours of the first cycle we have:

Hours of the second cycle= Total time in service- time Airworthiness Directive

=468-454=14

Then we have 14 hours of service and need to accumulate the rest to complete 454. It is

454-14=440

Then we need to accumulate 440 hours additional to complete the Airworthiness Directive .

6 0
3 years ago
This graph represents a transformation of the parent cube root function. Replace the values of h and k to create the equation of
Zigmanuir [339]

Answer: H is 5 and K is 2

Step-by-step explanation:

8 0
3 years ago
A suit with an original price of $290 was marked down 33%. You have a coupon good for an additional 36% off. What is the final c
TiliK225 [7]

Answer:

$20.01

Step-by-step explanation:

x 69

------ × --------

290 100

69 = 33 + 36

multiply 290 ×69 then divide by 100 to find c

4 0
3 years ago
Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
How do you simplify 220/1 x 5/2
Dafna11 [192]
The simplify answer is 550.
7 0
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