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aliina [53]
3 years ago
12

PLEASE HELP!!!! i will give branliest

Mathematics
1 answer:
borishaifa [10]3 years ago
7 0

Answer:The last answer doesn't have a solution of -2

Step-by-step explanation:

You might be interested in
Samantha drew Triangle JKL with the coordinates (2,3), (4, 3) and (5,2). She reflected this image over the x-axis to create an i
zavuch27 [327]

Answer:

J'(2,-3), K'(4, -3) and L'(5,-2)

Step-by-step explanation:

The given triangle has vertices at  J(2,3), K(4, 3) and L(5,2).

The transformation rule for a reflection in the x-axis is:

(x,y)\to (x,-y)

We substitute the points to obtain the coordinates  of the image triangle  J'K'L'

J(2,3)\to J'(2,-3)

K(4,3)\to K'(4,-3)

L(5,2)\to L'(5,-2)

In other words, we negate the y-coordinates of triangle JKL to obtain the coordinates of J'K'L'

See attachment for graph.  

3 0
3 years ago
A= 5,000 (1+.002*10)
Aneli [31]
I think The answer is 5100
3 0
3 years ago
Please help IM SO CONFUSED!
rodikova [14]

Answer:

18

Step-by-step explanation:

You can see from the table that 2 bracelets are made every 30 minutes.

There are 9 intervals of 30 minutes each in 4.5 hours.

So, the number of bracelets that are made in 4.5 hours is

9x2=18

Another way to see this is to note from the table that 4 bracelets are made in 60 minute, which is 1 hour.  So, in 4 hours, 16 bracelets will be made. Also, from the table, 2 bracelets are made in 30 minutes, which is 1/2 hour.  So, adding up, 18 bracelets are made in 4.5 hours.

4 0
3 years ago
Simplify. a2 · a5 a3 A) a3 B) a4 C) a7 D) a10
castortr0y [4]

a^{10}, that is D

when multiplying like variables add the exponents

= a^{2+3+5} = a^{10}


5 0
3 years ago
Read 2 more answers
R = sec(θ) − 2cos(θ), where -π/2 < θ < π/2
Alex

Answer:

  y = (x/(1-x))√(1-x²)

Step-by-step explanation:

The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:

  x = r·cos(θ)

  y = r·sin(θ)

  x² +y² = r²

__

  r = sec(θ) -2cos(θ)

  r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)

  r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²

  (x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations

  x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor

  y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)

  \boxed{y=\dfrac{x\sqrt{1-x^2}}{1-x}}

__

The attached graph shows the equivalence of the polar and rectangular forms.

4 0
3 years ago
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