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3 years ago

Someone help me with this please

Mathematics

1 answer:

Fofino [41]3 years ago

8 0

Answer:

pretty sure it's B, its just really hard to see

Step-by-step explanation:

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PLEASE HELP JUST KINDA CONFUSED ON THIS ONE I WILL MARK AS BRAINLIEST

WARRIOR [948]

Answer:

1 milliliter is 20 drops.

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3 years ago

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Svetlanka [38]

The answer is option 2,

2x10^10

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3 years ago

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A log raft is moving downstream at a speed of 3 km/h. To insure that there are no river jams a supervisor travels along the raft

skad [1K]

Answer:

2000 meters

Step-by-step explanation:

When the motorboat is moving downstream, its speed relative to the ground is 18 km/h, so its speed relative to the raft is 15 km/h.

When the motorboat is moving upstream, its speed relative to the ground is 12 km/h, so its speed relative to the raft is again 15 km/h.

Converted to m/min, the relative speed is:

15 km/h × (1000 m/km) × (1 h / 60 min) = 250 m/min

It takes the motorboat 16 minutes to travel to the front of the raft and back. Since the speed is the same in both directions, the motorboat takes 8 minutes to travel the length of the raft.

So the length of the raft is 250×8 = 2000 meters.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7Bb%20%2B%20%20-%20%20%5Csqrt%7B%7Bb%7D%5E%7B2%7D%20-%204ac%20%7D%20%7D%7

Phantasy [73]

If it is asking if that equation is the quadratic formula, then the answer is false. The reason why is that the first 'b' should be negative

The quadratic formula is

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

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago