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astraxan [27]
3 years ago
15

In 15–20, write the number positioned at each point.

Mathematics
1 answer:
Mumz [18]3 years ago
5 0

Answer:

See below.

Step-by-step explanation:

15.) -3.25

16.) -4.5

17.) 1.25

18.) -5.75

19.) 0.5

20.) -2.5

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s344n2d4d5 [400]

Answer:

Get a calculator a graphing calculator and insert each one of those in it

4 0
3 years ago
I need help like fast i am on 3 and i do not know what to do
Elis [28]

Answer:

Number 3 has no solution

Step-by-step explanation:

First you need to divide both sides of the equation 12x - 12y = -12 by 4

-3x + 3y = -3

12x - 12y = -12

-3x + 3y = -3

3x - 3y = -3

Then you need to eliminate one variable by adding the equations.

0 = -6

this statement is false, and therefore has no solution.

7 0
3 years ago
R(x) = (500 + ax) (50 - bx)
o-na [289]

Answer:

x

=

−

500

b

−

50

a

+

R

±

√

250000

b

2

+

2500

a

2

−

100

a

R

+

1000

b

R

+

R

2

+

50000

a

b

2

a

b

Step-by-step explanation:

4 0
3 years ago
Find what h equals please help ASAP THANK YOU!<br><br> 4h-1=3h+2
Sunny_sXe [5.5K]
H = 3.

FIRST STEP:
<span>Add 1 to both sides to get rid of the -1 on the left side.
4h-1 = 3h+2
</span><span>4h-1 (+1) = 3h+2 (+1)
</span><span>4h = 3h+3
 
SECOND (FINAL) STEP:
Subtract 3h from both sides to get rid of the 3h on the right side.
</span>4h(-3h) = 3h+3 (-3h)
h = 3
Hope this helps, sorry if it's hard to understand :)
3 0
3 years ago
Solve the Law of Cosine: c^2 = a^2+ b^2 - 2abcosC for cos C.
Andrew [12]

Answer:

The Law of Cosine :  cos C = \frac{a^{2}+ b^{2}-c^{2}}{2ab}

Step-by-step explanation:

See the figure to understand the proof :

Let A Triangle ABC with sides a,b,c,

Draw a perpendicular on base AC of height H meet at point D

Divide base length b as AD = x -b   and    CD = x

By Pythagoras Theorem

In Triangle BDC             And     In Triangle BDA

a² = h² + x²     (  1  )                        c² = h² + (x-b)²

                                                      c² = h² + x² + b² - 2xb   ...(. 2)

From above eq 1 and 2

c² = (a² - x²) + x² + b² - 2xb

or, c² = a² + b² - 2xb                    .....(3)

Again in ΔBDC

cos C = \frac{BD}{BC}

Or, cos C = \frac{x}{a}

∴ x= a cos C

Now put ht value of x in eq 3

I.e, c² = a² + b² - 2ab cos C

Hence , cos C = \frac{a^{2}+ b^{2}-c^{2}}{2ab}      Proved   Answer

6 0
3 years ago
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