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love history [14]
3 years ago
6

What is the answer to:3(3x-4)-2(2x-1)

Mathematics
2 answers:
KIM [24]3 years ago
7 0
The answer is x=-2 after you solve the equation
Tamiku [17]3 years ago
3 0

Answer:

= 5x -10

Step-by-step explanation:

=(3)(3x)+(3)(−4)+(−2)(2x)+(−2)(−1)

=9x+−12+−4x+2

Combine Like Terms:

=9x+−12+−4x+2

=(9x+−4x)+(−12+2)

=5x+−10

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2x + y = -4<br> 2x + 3y = 4
kirill115 [55]

Answer: x=4 ; y=<u> -4</u>

                            3

Step-by-step explanation:

2x + y = -4 .(-3)

2x + 3y = 4

-6x-3y=12

<u>2x + 3y = 4</u>

4x=16

x= <u>16</u>

    4

x=4

2x + 3y = 4

2.4+3y=4

8+3y=4

3y= -8+4

y=<u> -4</u>

   3

7 0
3 years ago
A sector of a circle has a central angle to 10° and arc length 28 pi units what is the radius of the circle
Dennis_Churaev [7]

Answer:

Radius = 504units

Step-by-step explanation:

Angle = 10°

Length of arc = 28π units

L = ∅/360 × 2πr

28π = 10/360 × 2πr

28π = 1/36 × 2πr

2πr = 28π × 36

π cancelled off.......

2r = 28 × 36

r = (28 × 36)/2

r = 28 × 18

r = 504units

8 0
3 years ago
How are the real solutions of a quadratic equation related to the graph of the quadratic function?
Anika [276]
The solutions you get when you solve the formula are the corresponding y coordinates to your x value. So say a point on your graph is (2,3). The first number is x and the second is y. (x,y). The number you plug into your function is x,or in this case: 2. The solution to the equation when the x value is plugged in is y, or 3. Therefore, giving you a point on your graph.
7 0
3 years ago
Find k so that the distance from (–1, 1) to (2, k) is 5 units. k= k= *there are two solutions for 2*
dalvyx [7]

Answer:

k = -3

k =5

Step-by-step explanation:

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\d = 5\\(-1,1) =(x_1,y_1)\\(2,k)=(x_2,y_2)\\

5=\sqrt{\left(2-\left(-1\right)\right)^2+\left(k-1\right)^2}\\\\\mathrm{Square\:both\:sides}:\quad 25=k^2-2k+10\\25=k^2-2k+10\\\\\mathrm{Solve\:}\:25=k^2-2k+10:\\k^2-2k+10=25\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\k^2-2k+10-25=25-25\\k^2-2k-15=0\\\\\mathrm{Solve\:by\:factoring}\\\\\mathrm{Factor\:}k^2-2k-15:\quad \left(k+3\right)\left(k-5\right)\\\mathrm{Solve\:}\:k+3=0:\quad k=-3\\

\mathrm{Solve\:}\:k-5=0:\quad k=5\\\\k =5 , k=-3

7 0
3 years ago
Read 2 more answers
Solve. 3t &lt; –15 (1 point)<br> t &lt; –5<br> t &gt; –5<br> t &lt; –45<br> t &gt; –45
Kipish [7]
I hope this helps you



3t÷3 < -15÷3


t < -5
8 0
3 years ago
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