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2 years ago

A point has coordinates (-3,-3). Where is it located in the coordinate plane?

Mathematics

1 answer:

ludmilkaskok [199]2 years ago

5 0

C quadrant 3 because negative x value and negative y value

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If the surface area of a cube is increased by a factor of 2 (so that the new surface area is twice the size of the original surf

Mariana [72]

Answer:2.828

Step-by-step explanation:

Given

Surface area of a cube is increased by a factor of 2

Let the original surface area be A

such that new surface area is 2A

for volume we need to know new radius

$6(a')^2=2\times 6r^2$

where r'=new radius

r=original radius

$a'=\sqrt{2}a$

New volume(V') $=(a')^3=(\sqrt{2}a)^3$

$V'=2\sqrt{2}a^3$

Original volume $=a^3$

$\frac{V'}{V}=\frac{2\sqrt{2}a^3}{a^3}$

$V'=2\sqrt{2}V=2.828V$

8 0

2 years ago

I need help please and thanks

dimulka [17.4K]

Answer:

Step-by-step explanation:

n is the number of selections and k the number selected, that is

n = 5 and k = 2

note that n! = n(n - 1)(n - 2) ..... × 3 × 2 × 1, thus

$\frac{5(4)(3)(2)(1)}{2!(5-2)!}$

= $\frac{120}{2(1)3(2)(1)}$

= $\frac{120}{2(6)}$

= $\frac{120}{12}$

= 10

5 0

2 years ago

How do you write 3/6 in the simplest form

Molodets [167]

1/2 is the answer because 3 can divide into 6 evenly

4 0

2 years ago

Please help me solve the problem

Elina [12.6K]

Answer:

x = $\sqrt{2}$ ; y = 2

Step-by-step explanation:

You have a 45 -45 - 90 triangle. There are two things to know"

1. The hypothenuse = leg × $\sqrt{2}$

2. The legs are equal.

The hypothenuse is opposite the right angle that is "y" in this problem.

Hypothenuse = leg × $\sqrt{2}$

= $\sqrt{2} \sqrt{2} = \sqrt{4} = 2$

The legs are =

Therefore x = $\sqrt{2}$

7 0

2 years ago

Lucy has two star systems left to visit on her voyage, but her ship is running low on fuel. The first system, KA-77, is 12001200

-Dominant- [34]

Answer:

1348 light years.

Step-by-step explanation:

Please find the attachment.

Let x represent the distance between KA-7 and KA-11.

We have been given that the first system, KA-7, is 1200 light years away while the second system, KA-11, is 1700 light years away. Lucy sees an angle of 52 degrees between KA-7 and KA-11.

We can see from our attachment that Lucy, KA-7 and KA-11 forms a triangle and we will use law of cosines to solve for x.

$(AB)^2=(AC)^2+(BC)^2-2(AC)(BC)*cos (C)$

Upon substituting our given values in above formula we will get,

$x^2=1200^2+1700^2-2*1200*1700*cos (52^o)$

$x^2=1440000+2890000-4080000*0.615661475$

$x^2=4330000-2511898.81933008$

$x^2=1818101.18066992$

Let us take square root of both sides of our equation.

$x=\sqrt{1818101.18066992}$

$x=1348.3698234\approx 1348$

Therefore, KA-7 and KA-11 are approximately 1348 light years apart.

6 0

2 years ago