What is the area of the figure? Help quickly

Whole numbers greater than 6 but less than 16

Delvig [45]

7,8,9,10,11,12,13,14,15

4 0

3 years ago

3.The legs of a right triangle are equal. Given the length of the legs, find the length of the hypotenuse. Round

Zepler [3.9K]

Answer:

1.41a

Step-by-step explanation:

Let ABC is a right triangle with equal legs AB =BC.

Hence, ∠B is 90° and CA is the hypotenuse.

Now, applying Pythagoras Theorem, we can write

CA² = AB² + BC²

Let us assume that each legs AB = BC = a.

Therefore, CA² = a² +a² = 2a²

⇒Hypotenuse or CA = a√2 = 1.41a { Round to the nearest tenth} (Answer)

3 0

3 years ago

{WILL GIVE BRAINLIEST} pls solve show work or just do on paper and send me a picture

Nadusha1986 [10]

Answer:

you torched a kid today

Step-by-step explanation:

i hope you are happy:(

3 0

3 years ago

Help me plz Use the distance formula to find the distance between each pair of points Round to the nearest tenth if necessary nu

irakobra [83]

D=√(x2-x1)² + (y2-y1)²
D=√(-6 3/4-8 1/2)² + (7 1/2-12)²
D=√(-15 1/4)² + (-4 1/2)²
D=√3721/16 + 81/4
D=√4069/16
D=64/4
D=16

4 0

4 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$

Then

$\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.$

Answer 2: correct choice is D.

8 0

4 years ago

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What is the area of the figure? Help quickly​

What is the area of the figure? Help quickly