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Kryger [21]
3 years ago
5

If someone is 5’10 and weights 119 pounds what is their BMI? Show work

Mathematics
1 answer:
Korolek [52]3 years ago
6 0

hello,

if i'm 5'10 and weigh 119 pounds this includes that my BMI: 17.07

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350 cm^2 (A for connections academy)
3 0
4 years ago
A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =
tia_tia [17]

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

f(x)=-16x^2+64x+5

Where f(x) is the height after x seconds.

We want to determine the time(s) when the ball is 10 feet in the air.

Therefore, we will set the function equal to 10 and solve for x:

10=-16x^2+64x+5

Subtracting 10 from both sides:

-16x^2+64x-5=0

For simplicity, divide both sides by -1:

16x^2-64x+5=0

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Substitute:

\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}

Evaluate:

\displaystyle x=\frac{64\pm\sqrt{3776}}{32}

Simplify the square root:

\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}

Therefore:

\displaystyle x=\frac{64\pm8\sqrt{59}}{32}

Simplify:

\displaystyle x=\frac{8\pm\sqrt{59}}{4}

Approximate:

\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

7 0
3 years ago
Can you help me with this?
horsena [70]

Answer:

B

Step-by-step explanation:

it this answer

7 0
3 years ago
Do alto de um farol cuja altura é de 20 m avista se um navio sob ângulo de depressão de 30°. A que distância , aproximadamente,
Mkey [24]
Esta é a trigonometria . Se você desenhar uma linha a partir do topo da casa de luz para o barco, você terá a hypotonuse de um triângulo. Um truque é lembrar que este é um triângulo especial. É um triângulo 30-60-90 , que tem propriedades especiais mostradas na fixação abaixo . por isso sabemos que o lado adjacente que não é o hyposonuse é x√3 . Agora sabemos que x<span>√3 = 20
Solve for x.
x</span><span>√3=20
divide both sides by </span><span>√3.
x=20/(</span><span><span>√3)
</span>Try not to have square roots (</span><span><span>√)</span> in denomenator so multiply top and bottom by </span><span>√3 and get
x=(20</span><span>√3)/3
x is what we are looking for so the answer is </span>
20<span>√3 m </span><span>ou cerca de 34.64 m</span>










7 0
4 years ago
15 points, no messing around, please do it right
cupoosta [38]

Answer:

416 is your answer!!!

Step-by-step explanation:

mak me brainliest and add me as a fiend

8 0
3 years ago
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