All categories

3 years ago

Pls can any of you help with this.

Mathematics

1 answer:

Lorico [155]3 years ago

3 0

X=6 and Y= 6 square root 2

You might be interested in

Question 1

vivado [14]

Answer:

33, 792

Step-by-step explanation:

first convert the 8 miles to feet because our answers are in feet.

1 mile= 5280 feet

therefore 8 miles= 5280×8

= 42240

Dale and son covered 4/5 of the 42240 feet

= 4/5×42240

= 33,782 feet

4 0

2 years ago

Give me an expression of nine less than a number

kompoz [17]

So if you want it to be 9 less than your number (x) then you would subtract nine.

So it would be:

x - 9.

So if your "number" is 12 then you would substitute x for 12 and it would be 12-9. The answer is 3. So 3 is 9 less than a number (12).

I believe

4 0

3 years ago

PLZ HELP ASAP!!!!!!!!!!!!!!!!!!!!!!

nika2105 [10]

2. A
4. D
I don't know the rest but good luck.

4 0

3 years ago

All sets that contain the number 5 ?

vivado [14]

Real numbers, rational numbers, Integers, Whole numbers, and Natural numbers.

5 0

3 years ago

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th

wel

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$