All categories

2 years ago

I need help correct gets crown answer quick!

Mathematics

2 answers:

Ne4ueva [31]2 years ago

8 0

Answer:

6,7, and 10

Step-by-step explanation:

If this answer helped you then please mark it as brainliest so I can get to the next rank

Helen [10]2 years ago

5 0

6,7,11

Hope that helps!!

You might be interested in

If the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2:3

horrorfan [7]

Answer: The correct option is (B) 24 : 25.

Step-by-step explanation: Given that the perimeter of square region S and the perimeter of rectangular region R are equal and the sides of R are in the ratio 2 : 3.

We are to find the ratio of the area of R to the area of S.

Let 2x, 3x be the sides of rectangle R and y be the side of square S.

Then, according to the given information, we have

$\textup{Perimeter of rectangle R}=\textup{Perimeter of square S}\\\\\Rightarrow2(2x+3x)=2(y+y)\\\\\Rightarrow 5x=2y\\\\\Rightarrow \dfrac{x}{y}=\dfrac{2}{5}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Therefore, the ratio of the area of R to the area of S is

$\dfrac{2x\times3x}{y\times y}\\\\\\=\dfrac{5x^2}{y^2}\\\\\\=6\left(\dfrac{x}{y}\right)^2\\\\\\=6\times\left(\dfrac{2}{5}\right)^2~~~~~~~~~~~[\textup{Using equation (i)}]\\\\\\=\dfrac{24}{25}\\\\=24:25.$

Thus, the required ratio of the area of R to the area of S is 24 : 25.

Option (B) is CORRECT.

6 0

3 years ago

Given the Graph, identify the y-intercept

loris [4]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

5th grade math. correct answer will be marked brainliest, select which is greater than,equal to, less than for all

lozanna [386]

Answer:

Step-by-step explanation:

1. Less than

2. Greater than

3. Equal to

4. Greater than

5. Equal to

6. Less than

4 0

3 years ago

3/4 + 2/3 = 1 5/12 eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

satela [25.4K]

3/4+2/3= 17/12

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

8. Find all the real fourth roots of 256

mezya [45]

X^4=256
x^4-256=0
(x^2)^2-(16)^2=0
(x^2+16)(x^2-16)=0
either x^2+16=0
it gives complex roots.
or x^2-16=0
or x^2-4^2=0
(x+4)(x-4)=0
either x+4=0,x=-4
or x-4=0
x=4

7 0

3 years ago