All categories

3 years ago

Each side length is doubled. What is the PERIMETER of the new figure?

Mathematics

2 answers:

lesantik [10]3 years ago

5 0

Answer:

24 units

Step-by-step explanation:

Original perimeter:

5+4+3=12

When each side length is doubled, the original perimeter is doubled.

12*2=24

nignag [31]3 years ago

4 0

Answer:

if the length of each side is is doubled then

5×2 = 10

4×2 = 8

3×2 = 6

perimeter of new figure = 10 + 8 + 6

= 24 units

hope this helps good day mate

You might be interested in

10. Shown is a circle centre o

Aleksandr [31]

Answer:

Step-by-step explanation:

6 0

3 years ago

If you if you take away 25% of the radius of the suspension how much raise would you have left over of the hundred percent you t

natulia [17]

Answer:

Step-by-step explanation:

25% of 100 is 25

100 - 25 = 75

Give me brainllest

3 0

3 years ago

Help!!!!!!!Plz!!!!!!!

Hoochie [10]

Answer:

I'm so confused

Step-by-step explanation:

6 0

3 years ago

A cubic function with roots at 2, 0 and 3 containing the point at (5,-6)

natulia [17]

Answer:

Cubic equations and the nature of their roots

Just as a quadratic equation may have two real roots, so a cubic equation has possibly three. But unlike a quadratic equation which may have no real solution, a cubic equation always has at least one real root.

5 0

2 years ago

If m and n are the zeroes of the polynomial 6y 2 -7y – 2, find a polynomial whose zeroes are 1/m and 1/n

denis-greek [22]

Answer:

$y^2-\frac{7}{2}y+3$

Step-by-step explanation:

Given polynomial is,

6y² - 7y - 2

Fro the zeros of the given polynomial,

6y² - 7y - 2 = 0

6y² - 4y - 3y - 2 = 0

2y(3y - 2) - 1(3y - 2) = 0

(2y - 1)(3y - 2) = 0

(2y - 1) = 0

y = $\frac{1}{2}$

(3y - 2) = 0

y = $\frac{2}{3}$

Therefore, zeros of this polynomial are m = $\frac{1}{2}$ and n = $\frac{2}{3}$

If a polynomial has zeros as $\frac{1}{m}$ and $\frac{1}{n}$ then the zeros will be $\frac{3}{2}$ and 2.

Polynomial will be,

$(y-\frac{3}{2})(y-2)$

= $y(y-2)-\frac{3}{2}(y-2)$

= $y^2-2y-\frac{3}{2}y+3$

= $y^2-\frac{7}{2}y+3$

7 0

2 years ago