2 years ago

If the area of figure 1 is 6 and figure 2 is 24. figure 3 is 54 then what area for figure 4 and 5

Mathematics

1 answer:

Nitella [24]2 years ago

7 0

2+2 is 4 minus 1 thats 3 quick maths

You might be interested in

Find the area of the parallelogram

BartSMP [9]

6*14=84
the answer is 84cm²

7 0

3 years ago

Read 2 more answers

What’s 3 times 6 over 7 **MULTIPLY

Amanda [17]

Would it be 2.5??? I think it would but not completely sure

7 0

2 years ago

Read 2 more answers

What's -5x – 10 = 10

Nikitich [7]

-5x-10=10 x= -4

Work:
-5x-10=10
+10 +10
______________
-5x=20
_______
-5 -5

x=-4

3 0

2 years ago

Read 2 more answers

Find the volume for the regular pyramid.<br><br><br><br> V =

Paul [167]

Answer:

What do you mean by that?

4 0

2 years ago

In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

vichka [17]

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$

So, $x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

3 0

2 years ago

Read 2 more answers