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2 years ago

Find the value of x.

Mathematics

1 answer:

VikaD [51]2 years ago

4 0

Answer:

X = 9

okay If I'm wrong sry what's 9x3 it's 45 right ( me not using a caluctalter ) ( I didn't even spell that Right lol ) but anyways it's X=9

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Hilda's personal information is shown below. According to the following table what is her hypothetical

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606

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2 years ago

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A butterfly is 12 inches long and 13 inches wide. Lashay makes a scale drawing

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Answer:

1.08 inches

Step-by-step explanation:

13 / 12 , to keep aspect ratio the same

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1 year ago

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Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

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2 years ago

an NFL team kicker misses 2 out of every 11 field goals if he missed 8 filed goals throughout the season how many did he attempt

Aleks [24]

For this equation, you want to do it in fractions/ratios to properly solve it. You would have his average misses out of every field goal and his real missed attempts over total. It would look like this

$\frac{2}{11}$ = $\frac{8}{x}$

You want to solve for x since x is the total amount of field goals that he attempted. You can do this by doing cross multiplication:

(2)(x) = (8)(11)

From here you can get:

2x = 88

Divide each side by 2 to isolate x and you get:

x= 44

So he made a total of 44 field goals.

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Five buses leave on a field trip .There are about 45 students per bus.About how many students are on the 5 buses.

dangina [55]

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