4x 2 is less than or equal to -32

Answers

Answer 1

Answer: Less than is the answer

Answer 2

Answer: less than is the answer

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Which radical expression is equivalent to 3^3/4

Answers

The numerator of the fraction is the power of the number and the denominator is the root of the number. This is expressed as follows:

⁴√(3)³

Answer:27

Step-by-step explanation:

Solve 2(x + 1) = 2x + 2.

Answers

2(x+1)=2x+2
^^^^^^2 gets distributed to both sides in the brackets.
the answer comes out to 2x+2.
therefore, 2x+2=2x+2.

Can someone please answer this I’m stuck

Answers

The system has no solutions.
To do this question I started out by trying to solve the simultaneous equations using substitution.
First we can simplify 6x = 12y - 24 by dividing everything by 6, which gives us:
x = 2y - 4
Then we can substitute this into the second equation:
30y - 15(2y - 4) + 30 = 0
30y - 30y + 60 + 30 = 0
90 = 0
As this is obviously not true, we can conclude that the system has no solution.
I hope this helps! Let me know if you have any questions :)

What is the prime factorization of 1,260? A. 2 × 2 × 3 × 3 × 5 × 7
B. 4 × 5 × 7 × 9
C. 2 × 3 × 5 × 6 × 7
D. 2 × 3 × 5 × 7

Answers

A. 2*2*3*3*5*7

Whilst B and C both also yield 1,260, they aren't broken down in to their prime factors. In B, 4 can be broken down in to 2*2 and 9 in to 3*3, and in C, 6 can be broken down in to 2*3

And D just doesn't total 1,260

Answer:

A.2x2x3x3x5x7

Step-by-step explanation:

Penn foster

What is the area of this figure?

Answers

Answer:

35 in^2

Step-by-step explanation:

15+20=35

PLEASE HELP. What is the equation of the circle in standard form where the center is (-16,-14) and another point on the circle is (-8,-8)

Answers

recall that the radius is the distance from the center of a circle to any point on the circle.

we know the center is at -16,-14, and we know that -8,-8 is a point on the circle, so the distance between both must be the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\n\n(\stackrel{x_1}{-16}~,~\stackrel{y_1}{-14})\qquad(\stackrel{x_2}{-8}~,~\stackrel{y_2}{-8})\qquad \qquadd = √(( x_2- x_1)^2 + ( y_2- y_1)^2)\n\n\n\stackrel{radius}{r}=√([-8-(-16)]^2+[-8-(-14)]^2)\n\n\nr=√((-8+16)^2+(-8+14)^2)\implies r=√(8^2+6^2)\n\n\nr=√(100)\implies \boxed{r=10}\n\n[-0.35em]\rule{34em}{0.25pt}$

$\bf \textit{equation of a circle}\n\n(x- h)^2+(y- k)^2= r^2\qquadcenter~~(\stackrel{-16}{ h},\stackrel{-14}{ k})\qquad \qquadradius=\stackrel{10}{ r}\n\n\n\[x-(-16)]^2+[y-(-14)]^2=10^2\implies \blacktriangleright (x+16)^2+(y+14)^2=100 \blacktriangleleft$

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