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dimaraw [331]
3 years ago
7

Help me please number 7 (please show work)

Mathematics
1 answer:
astra-53 [7]3 years ago
8 0

Answer:

m<H=30 degrees by the Triangle Sum Theorem, so <H=<V. <G=<U, they are right angles. If GH=UV, triangle FGH=triangle TUV by the ASA.

x=3  and y=8

Step-by-step explanation:

H equals 30 by 180-150=30 (getting 150 by adding 60 and 90)

I know this is not much of a answer, but I hope it helps a little.

http://www.somersetkey.com/ourpages/auto/2017/12/13/63309824/geo%20ch%205%20sec%204%20teach.pdf

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What is the answer to the problem
andreyandreev [35.5K]
1.5x7 = 10.5 cups per week

I don't know how to convert it. probably a decimal or fraction.

here's how to convert, 10.5/16 = # gallon(s)
7 0
2 years ago
How to factor x^2-2x-15
olga2289 [7]

Answer:

(x - 5)(x + 3)

Step-by-step explanation:

Given

x² - 2x - 15

Consider the factors of the constant term (- 15) which sum to give the coefficient of the x- term (- 2)

The factors are - 5 and + 3, since

- 5 × 3 = - 15 and - 5 + 3 = - 2, thus

(x - 5)(x + 3) ← in factored form

6 0
3 years ago
A student throws a rock into the sky from the top of the Webster build- ing with an angle α = π 4 from the horizontal line and t
lord [1]

Answer:

  • r(0) = <0, 100> . . . . . . . .meters
  • r'(0) = <7.071, 7.071> . . . . meters per second

Step-by-step explanation:

<u>Initial Position</u>

The problem statement tells us we're measuring position from the ground at the base of the building where the projectile was launched. The initial horizontal position is presumed to be zero. The initial vertical position is said to be 100 meters from the ground, so (in meters) ...

  r(0) = <0, 100>

<u>Initial Velocity</u>

The velocity vector resolves into components in the horizontal direction and the vertical direction. For angle α from the horizontal, the horizontal component of velocity is v₁·cos(α), and the vertical component is v₁·sin(α). For v₁ = 10 m/s and α = π/4, the initial velocity vector (in m/s) is ...

  r'(0) = <10·cos(π/4), 10·sin(π/4)>

  r'(0) ≈ <7.071, 7.071>

7 0
3 years ago
X+y+z=12<br> 6x-2y+z=16<br> 3x+4y+2z=28<br> What does x, y, and z equal?
lianna [129]

Answer:

x = 20/13 , y = 16/13 , z = 120/13

Step-by-step explanation:

Solve the following system:

{x + y + z = 12 | (equation 1)

6 x - 2 y + z = 16 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Swap equation 1 with equation 2:

{6 x - 2 y + z = 16 | (equation 1)

x + y + z = 12 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/6 × (equation 1) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+(4 y)/3 + (5 z)/6 = 28/3 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Multiply equation 2 by 6:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/2 × (equation 1) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+5 y + (3 z)/2 = 20 | (equation 3)

Multiply equation 3 by 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+10 y + 3 z = 40 | (equation 3)

Swap equation 2 with equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+8 y + 5 z = 56 | (equation 3)

Subtract 4/5 × (equation 2) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+(13 z)/5 = 24 | (equation 3)

Multiply equation 3 by 5:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+13 z = 120 | (equation 3)

Divide equation 3 by 13:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract 3 × (equation 3) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y+0 z = 160/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 2 by 10:

{6 x - 2 y + z = 16 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Add 2 × (equation 2) to equation 1:

{6 x + 0 y+z = 240/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract equation 3 from equation 1:

{6 x+0 y+0 z = 120/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 1 by 6:

{x+0 y+0 z = 20/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Collect results:

Answer:  {x = 20/13 , y = 16/13 , z = 120/13

8 0
3 years ago
Fill in the table using this function rule
marusya05 [52]
<h3>hello!</h3>

Plug in the values of x and solve for y:-

y=5+4x

y=5+4(4)

y=5+16

y=21

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y=5+4x

y=5+5(5)

y=5+20

y=25

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y=5+4x

y=5+4(8)

y=5+32

y=37

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y=5+4x

y=5+4(10)

y=5+40

y=45

__________________________________

<h3>note:-</h3>

Hope everything is clear; if you need any clarification/explanation, kindly let me know, and I will comment and/or edit my answer :)

8 0
2 years ago
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