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Dmitriy789 [7]
3 years ago
5

Find the gcf of 12,30,72

Mathematics
1 answer:
kvv77 [185]3 years ago
7 0

Answer: 6


Step-by-step explanation:

12 = 2^2 • 3

30 = 2 • 3 • 5

72 = 2^3 • 3^2


You might be interested in
Which of the following are roots of the following polynomial x^3-11x^2+33x+45
Mandarinka [93]
For x^3-11x^2+33x+45 , we can make it an equation so <span>x^3-11x^2+33x+45=0. Next, we can find out if -1 or -3 is a factor. If -1 is a factor, than (x+1) is factorable. Using synthetic division, we get 

      x^2-12x+45
       ___ ________________________
x+1 | x^3-11x^2+33x+45

      - (x^3+x^2)
       _________________________
            -12x^2+33x+45
          - (-12x^2-12x)
          ______________
            45x+45
         -(45x+45)
___________
0


Since that works, it's either B or D. We just have to figure out when
</span> x^2-12x+45 equals 0, since there are 3 roots and we already found one. Using the quadratic formula, we end up getting (12+-sqrt(144-180))/2=
(12+-sqrt (-36))/2. Since sqrt(-36) is 6i, and 6i/2=3i, it's pretty clear that B is our answer
8 0
3 years ago
Please help! I cannot figure this out, I have no idea what to do.
joja [24]

3. First you should identify which line corresponds to which equation.

x + 2y = -8   ⇒   y = -x/2 - 4

is the line with slope -1/2 and intercept (0, -4), so the first inequality refers to the lower line in the graph.

For the inequality itself, the solution set that satisfies it is either the region above or below it. To decide which, pick any point in the plane and plug its coordinates into the inequality. The origin is a natural choice, since it's not on the line and working with 0s is easy.

In the first inequality, we have

x = 0 and y = 0   ⇒   0 + 2•0 = 0 < -8

which is not true. So (0, 0) is not in the solution set, and it stands to reason that any point in the same region will not be a solution. In other words, the region above the line x + 2y = -8 does not satisfy the inequality, while the one below it does.

In the second inequality,

x = 0 and y = 0   ⇒   2•0 + 0 = 0 ≥ -1

which is true. This means the region containing (0, 0) above the upper line solves the second inequality.

The solution to the system of inequalities is then the intersection of these two regions. (See attached; I've labeled the solution to the first inequality in blue, the solution to the second one in red, and the solution to both in green.)

All this work is kind of overkill for this particular question, though. All you really need to do is check if the given point satisfies the inequalities:

• (-5, 5) :

x = -5 and y = 5   ⇒   -5 + 2•5 = 5 < -8   ⇒   NO

• (2, -5) :

x = 2 and y = -5   ⇒   -2 + 2•5 = 8 < -8   ⇒   NO

• (-4, -6) :

x = -4 and y = -6   ⇒   -4 + 2•(-6) = -16 < -8   ⇒   MAYBE

x = -4 and y = -6   ⇒   2•(-4) + (-6) = -14 < -8   ⇒   YES

• (5, -7) :

x = 5 and y = -7   ⇒   5 + 2•(-7) = -9 < -8   ⇒   MAYBE

x = 5 and y = -7   ⇒   2•5 + (-7) = 3 < -8   ⇒   NO

4. x is the number of candy boxes with chocolates and y is the number of boxes without chocolates. Each of the x boxes cost $27.50, so if you have x boxes, their total cost is $27.5x. Each of the y boxes cost $25.00, so y boxes cost $25y. The sponsor doesn't want to order more than 100 boxes total, so

x + y < 100

but wants to raise at least $2000, so

27.5x + 25y ≤ 2000

Now do the same thing as before; plug in the listed x- and y-coordinates and pick the point that satisfies both inequalities. You will find that (50, 30) is the correct choice.

5. Consult the "overkill" part of problem 3. The line y = -3x + 6 has a negative slope, so it's the downward sloping one. Check if the origin satisfies the inequality:

x = 0 and y = 0   ⇒   0 ≤ -3•0 + 6   ⇒   0 ≤ 6   ⇒   YES

This means Regions II and III solve the first inequality.

Do the same with the other inequality:

x = 0 and y = 0   ⇒   0 ≥ 1/2•0 + 1   ⇒   0 ≥ 1   ⇒   NO

This tells us that Regions I and II solve the second inequality.

The intersection of these regions is of course Region II.

6. One of the plotted lines is apparently y = x + 4, which has a positive slope, so this must be the upper line. The shaded region below it corresponds to the solution of either y < x + 4 or y ≤ x + 4 and we eliminate B.

The other line has negative slope, which eliminates A and D since

3x - 4y = 20   ⇒   y = 3/4x - 5

has positive slope. This leaves D.

6 0
2 years ago
To eliminate the terms and solve for y in the tewest steps, by which constants should the equations be multiplied by before
Novosadov [1.4K]

Answer:

the equation is not shown...

Step-by-step explanation:

but to solve for y you should use the first equation that is use the coefficient of x in the first equation to multiply all the terms in the second equation.

And use the coeffient of x in the second equation to multiply the the terms in the first equation.

It would have been better if the equation was shown for better explanation.

6 0
3 years ago
The park is 6 mi due east of Freeport High School, and the library is 4 mi due west of the school. The police department is loca
AURORKA [14]
The answer is 6.7 mi.

Since the right triangle is present, to calculate this we will use the Pythagorean theorem. According to the Pythagorean theorem, the square of the hypotenuse (c²) is equal to the sum of the squares of two other sides (a² + b²)<span>:
c</span>² = a² + b²<span>

First, we need to express symbols:
b - side of the right triangle (distance </span><span>between police department and Freeport High School).
a1 - </span>side of the first right triangle (distance between library and Freeport High School).<span>
a2 - </span><span>side of the second right triangle (distance between park and Freeport High School).
c1 - hypotenuse of the first triangle (distance between </span>police<span> department and library).
c2 - </span>hypotenuse of the second triangle (distance between police<span> department and park).

Therefore, we need to calculate c2.
It is given:
a2 = 6 mi
a1 = 4 mi
c1 = 5 mi

First, let's calculate b, which is a common side for two triangles:
</span>c1<span>² = a1² + b²
</span>b² = c1<span>² - a1²
</span>b² = 5<span>² - 4²
</span>b² = 25<span> - 16
</span>b² = 9
√b² = √9
b = 3.

We know b, now we can calculate c2:
c2<span>² = a2² + b²
</span>c2<span>² = 6² + 3²
</span>c2<span>² = 36 + 9
</span>c2<span>² = 45
</span>√c2<span>² = </span>√45
c2 = 6.7

The distance between police department and park is 6.7 miles.

3 0
3 years ago
(06.04)
ira [324]

Answer:

itsw b

Step-by-step explanation:

4 0
3 years ago
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