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3 years ago

Can somebody please help me with this question?

Mathematics

1 answer:

kompoz [17]3 years ago

4 0

Answer:

Step-by-step explanation:

Solve the first half of the equation-

-8x+3 >/=27

-3 -3

-8x>/=24

divided by -8 on both sides

x</=-3

Now for the second side:

-13x+5>/= 57

-5 -5

-13x>/=52

divide by -13 in both sides

x</=-4

We now have two equations. The answer would be x</=-3 because -4 is also less than three, so it can be simplified down to -3. I know you might not have needed the explanation since it’s just khan academy, but I figured I’d explain it anyway :)

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Point R lies on the directed line segment from L (-8,-10) to M (4,-2) and partitions the segment in the ratio 3 to 5. What are t

Dvinal [7]

Answer:

R = (- 3.5, - 7 )

Step-by-step explanation:

Using the Section formula

$x_{R}$ = $\frac{3(4)+5(-8)}{3+5}$ = $\frac{12-40}{8}$ = $\frac{-28}{8}$ = - 3.5

$y_{R}$ = $\frac{3(-2)+5(-10)}{3+5}$ = $\frac{-6-50}{8}$ = $\frac{-56}{8}$ = - 7

Thus coordinates of R = (- 3.5, - 7 )

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3 years ago

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In the diagram, point O is the center of the circle. If m∠ABC = 70°, what is m∠AOC?

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3 years ago

Find the mode of this set of data 18,27,28,44,44,50,67

8_murik_8 [283]

Answer:

Step-by-step explanation:

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2 years ago

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Can someone plz help me with this!

asambeis [7]

Answer:

(arranged from top to bottom)

System #3, where x=6

System #1, where x=4

System #7, where x=3

System #5, where x=2

System #2, where x=1

Step-by-step explanation:

System #1: x=4

$2x+y=10\\x-3y=-2$

To solve, start by isolating your first equation for y.

$2x+y=10\\y=-2x+10$

Now, plug this value of y into your second equation.

$x-3(-2x+10)=-2\\x+6x-30=-2\\7x=28\\x=4$

System #2: x=1

$x+2y=5\\2x+y=4$

Isolate your second equation for y.

$2x+y=4\\y=-2x+4$

Plug this value of y into your first equation.

$x+2(-2x+4)=5\\x+(-4x)+8=5\\x-4x+8=5\\-3x=-3\\x=1$

System #3: x=6

$5x+y=33\\x=18-4y$

Isolate your first equation for y.

$5x+y=33\\y=-5x+33$

Plug this value of y into your second equation.

$x=18-4(-5x+33)\\x=18+20x-132\\-19x=-114\\x=6$

System #4: all real numbers (not included in your diagram)

$y=13-2x\\8x+4y=52$

Plug your value of y into your second equation.

$8x+4(13-2x)=52\\8x+52-8x=52\\0=0$

all real numbers are solutions

System #5: x=2

$x+3y=5\\6x-y=11$

Isolate your second equation for y.

$6x-y=11\\-y=-6x+11\\y=6x-11$

Plug in your value of y to your first equation.

$x+3(6x-11)=5\\x+18x-33=5\\19x=38\\x=2$

System #6: no solution (not included in your diagram)

$2x+y=10\\-6x-3y=-2$

Isolate your first equation for y.

$2x+y=10\\y=-2x+10$

Plug your value of y into your second equation.

$-6x-3(-2x+10)=-2\\-6x+6x-30=-2\\-30=-2$

no solution

System #7: x=3

$y=10+x\\2x+3y=45$

Plug your value of y into your second equation.

$2x+3(10+x)=45\\2x+30+3x=45\\5x=15\\x=3$

3 0

2 years ago