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Misha Larkins [42]

3 years ago

5

14.371 value of 1 in decimal form

1 answer:

artcher [175]3 years ago

7 0

0.001. hope that helped

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Please help i need this as soon as possible

masya89 [10]

It's a quadratic so the parent function will be
$y = x ^{2}$
We can see that it has been shifted to the right 5 units, which gives us a new function
$y = (x - 5) ^{2}$
It has also been shifted up 9 units:
$y = (x - 5) ^{2} + 9$
But it has also been flipped across the y-axis which makes this the final change we need to do:
$y = - (x - 5) ^{2} + 9$

8 0

3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

or

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

or

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

Determine the Sin for K and P<br> Determine el pecado para Ky P<br> Tanluso<br> 14/<br> 48<br> 48

joja [24]

Step-by-step explanation:

sin∅ = opposite / hypotenuse

Answer:

sin(K) = 48/50

sin(P) = 14/50

4 0

3 years ago

Im confused right now. What equasion is represented by the area model? Format as 2x2

algol [13]

40x30=120x5=600x6=3600

Awnser:3600

8 0

2 years ago

Using Euler’s Formula (F + V = E + 2), find the number of vertices.

Usimov [2.4K]

I don't know euler's formula but I can guess that the one you gave is enough

F=18
V=?
E=33
so

18+?=33+2
18+?=35
minus 18 from both siddes
?=17

vertices is 17

7 0

3 years ago

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