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3 years ago

12

Which linear equality will not have a shared solution set with the graphed linear inequality? y > Two-fifthsx + 2 y Negative

two-fifthsx – 5 y < Five-halvesx + 2

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

The answer is B

Step-by-step explanation:

Im on Edg right now and I got the answer

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Please help.................

evablogger [386]

It usually works well to do what the instructions tell you. Multiply each diameter by 3.14.

a) 3.14×6 cm = 18.84 cm

b) [can't read the picture]

d) 3.14×9 in = 28.26 in

e) 3.14×11 mm = 34.54 mm

7 0

3 years ago

2 dots plots with number lines going from 0 to 10. Plot A has 0 dots above 0, 1, and 2, 1 above 3, 2 above 4, 2 above 5, 2 above

alexandr402 [8]

Answer:

First one:

Both the mean and median are greater for Plot A than for Plot B

Step-by-step explanation:

Set A:

Mean:

[1×10 + 2×7 + 2×6 + 2×5 + 2×4 + 1×3]/10

= 5.7

Median:

Median position: (10+1)/2 = 5.5th value

(5+6)/2

Median = 5.5

Set B:

Mean:

[1×7 + 3×6 + 3×5 + 2×4 + 1×3]/10

= 5.1

Median:

Median position: (10+1)/2 = 5.5th value

(5+5)/2

Median = 5

Mean: A is greater

Median: A is greater

8 0

3 years ago

Please help me I really need help

Oliga [24]

Answer:

Step-by-step explanation:

(1) 2x - 6y = -12

(2) x + 2y = 14

There is a -6y and a +2y. Since they have opposite signs, I'll try to eliminate the y terms. (That's my choice. There is more than one way to solve these.)

Multiply eq. (2) by 3:

3x + 6y = 42

Then add the result to eq. (1) to eliminate the y terms:

2x - 6y = -12

3x + 6y = 42

------------------

5x = 30, so x = 6

Now plug the value of x into eq. (2) and solve for y:

6 + 2y = 14

2y = 8

y = 4

Why did I use eq. (2) to solve for y? Because it's less work. I could have used eq. (1) instead:

2(6) - 6y = -12

12 - 6y = -12

-6y = -24

y = 4

More than one way to solve.

5 0

3 years ago

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

5 0

2 years ago

Can someone please explain this to me please? Thank you : )

Ber [7]

Answer:

7/25

Step-by-step explanation:

if 7% are defective and they order 25% of their lightbulbs from Baker that will be your answer.

6 0

3 years ago