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Papessa [141]
3 years ago
8

How many different three-digit numbers can be formed which contain two digits that are the same, and a third digit that is diffe

rent from the other two?
Mathematics
1 answer:
Vlad [161]3 years ago
3 0
Tricky question. What are you trying to figure out?
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2/5+x=1 What equals x?
kolbaska11 [484]

Answer:

Step-by-step explanation:

2/5 + x = 1

Multiply each term by 5

2 + 5x = 5

5x = 5 - 2

5x = 3

x = 3/5

8 0
3 years ago
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The area of a circle is 39.62cm2.<br> Find the length of the radius rounded to 2 DP.
Shkiper50 [21]

Using the formula for area of a circle:

Area = pi x radius^2

Replace area with the given value:

39.62 = pi x radius^2

Divide both sides by pi

Radius^2 = 39.62/pi

Radius^2 = 12.62

Solve for radius by taking the square root of both sides

Radius = sqrt(12.62)

Radius = 3.55 cm

4 0
2 years ago
I put a picture<br><br> —PLEASE HELP
Karo-lina-s [1.5K]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Base of a cube is a square, and in the given figure it's stated that the measure of each side is " x " ft

Therefore it's Area is equal to : -

  • side \times side

  • x \times x

  • {x}^{2}

Now, since it's Area is given as 1/4 ft², let's equate the Areas to find the measure of each side ~

that is ~

  • {x}^{2}  =  \dfrac{1}{4}

  • x =  \sqrt{ \dfrac{1}{2}  \times  \dfrac{1}{2} }

  • x =  \dfrac{1}{2}  \:  \: feet

It's time to find the volume of cube that is ~

  • (side) {}^{3}

So, let's plug the value of side length (1/2 ft)

Volume is equal to ~

  • {  \bigg(\dfrac{1}{2} \bigg) }^{3}

  • \dfrac{1}{8}  \:  \: ft {}^{3}

Hence, the required 1/8 or 0.125 ft³

I hope it helped ~

\mathrm{✌TeeNForeveR✌}

3 0
3 years ago
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What is the answer?<br> 82.7(blank)<br> - 5.59<br> ----------
Goshia [24]
The answer will be 77.11
3 0
3 years ago
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Angles A and B are corresponding angles. The measure of A is 135°. What is the measure of B?
KiRa [710]

Answer:

135 degrees

Step-by-step explanation:

corresponding angles are angles that are equal to each other

6 0
2 years ago
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