All categories

3 years ago

Solve the inequality-6p<12 Ah this is due by tomorrow

Mathematics

1 answer:

777dan777 [17]3 years ago

3 0

-6p < 12 Divide -6 on both sides to get "p" by itself

$\frac{-6p}{-6} < \frac{12}{-6}$

p > -2

When you divide or multiply a negative number to both sides of the inequality, the sign (< , > , ≤ , ≥) is flipped.

You might be interested in

Given the equation, what is the center and radius of the circle? <br> (X-2) ^2+ (Y+11)^2=100

LekaFEV [45]

Answer:

Center: (2, -11)

Radius: 10

Step-by-step explanation:

(x - 2)² + (y + 11)² = 100

(x - h)² + (y - k)² = r²

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

x - 2 = 0

+2 +2

x = 2

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

y + 11 = 0

-11 -11

y = -11

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

r² = 100

r = √100

r = 10

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

Center: (h, k)

Radius: r

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

Center: (2, -11)

Radius: 10

I hope this helps!

3 0

1 year ago

Read 2 more answers

Muchii unui cub are lungimea de 16cm.Care este perimetrul unei freţe a cubului? va rog rapid dau coroana

MAVERICK [17]

Răspuns:

64 cm

Explicație pas cu pas:

Perimetrul fiecărei margini a unui cub:

4 * lungime laterală

Lungime laterală = 16cm

Prin urmare, perimetrul va fi:

4 * 16cm

= 64cm

6 0

2 years ago

Can someone show what Triggernometry looks like

elena-14-01-66 [18.8K]

Answer:

search it up in google

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

6000x700000 help please

Nadya [2.5K]

4.2e9!!!!!!!!!!!!!!!!!!!

7 0

3 years ago

Read 2 more answers

Find all solutions to the equation (x² + 5x + 6)(x² − 3x − 4) = 0.

Soloha48 [4]

Answer:

The solutions are:

$x=-2,\:x=-3,\:x=-1,\:x=4$

Step-by-step explanation:

To find the solutions to the equation $\left(x^2+5x+6\right)\left(x^2-3x-4\right)=0$ you need to:

Factor $\left(x^2+5x+6\right)$

Break the expression into groups

$x^2+5x+6=\left(x^2+2x\right)+\left(3x+6\right)$

Factor out $x$ from $(x^2+2x)$

$x^2+2x=x\left(x+2\right)$

Factor out 3 from $3x+6$

$3x+6=3\left(x+2\right)$

$x^2+5x+6=x\left(x+2\right)+3\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\\\x^2+5x+6=\left(x+2\right)\left(x+3\right)$

Factor $\left(x^2-3x-4\right)$

$x^2-3x-4=\left(x^2+x\right)+\left(-4x-4\right)\\\\x^2-3x-4=x\left(x+1\right)-4\left(x+1\right)\\\\x^2-3x-4=\left(x+1\right)\left(x-4\right)$

Therefore

$\left(x^2+5x+6\right)\left(x^2-3x-4\right)=\left(x+2\right)\left(x+3\right)\left(x+1\right)\left(x-4\right)=0$

Using the Zero Factor Theorem:

5 0

3 years ago