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1 year ago

Anyone know how to prove these are equal to one another? I also need all the steps.

Mathematics

1 answer:

IrinaVladis [17]1 year ago

6 0

Answer:

Step-by-step explanation:

it can be simplified to $\frac{sec \theta}{cot\theta+tan\theta} =sin\theta$ . if you multiply each side by the denominator, it becomes $sec\theta=\frac{sin^2\theta}{cos\theta} + cos\theta$ since $tan\theta=\frac{sin\theta}{cos\theta}$ and cotan is the reciprocal of that. Because $sec\theta=\frac{1}{cos\theta}$ so multiplying each side gives $1=sin^2\theta+cos^2\theta$ , and $sin^2\theta+cos^2\theta$ is equal to 1.

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When the sun has risen 32 degrees above the horizon, Sandy casts a shadow that is 9 feet 2 inches long. How tall is Sandy, to th

Makovka662 [10]

Answer:

The height of Sandy is 68.74 inches.

Step-by-step explanation:

It is given that the sun has risen 32 degrees above the horizon. It means the angle of elevation is 32°.

Let the height of Sandy be x.

Sandy casts a shadow that is 9 feet 2 inches long.

We know that 1 feet = 12 inches

Using this conversion, we get

Shadow = 9 feet 2 inches = 9(12)+2 = 110 inches

In a right angled triangle,

$\tan\theta =\frac{perpendicular}{base}$

In triangle ABC,

$\tan(\angle C)=\frac{AB}{BC}$

$\tan(32^{\circ})=\frac{x}{110}$

Multiply both sides by 110.

$110\tan(32^{\circ})=x$

$68.73562871=x$

$x\approx 68.74$

Therefore the height of Sandy is 68.74 inches.

8 0

3 years ago

PLEASE HELP!!!!

Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

if "a" is positive, then opens UP
if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

the Axis Of Symmetry (x-value) is: x = $\frac{-b}{2a}$
y-value is found by plugging in the AOS for "x" in the equation

****************************************************************************************

1) y = x² + 11x + 24

a = +1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$

y = $(-\frac{11}{2})^{2}$ + 11 $(-\frac{11}{2} )$ + 24 = $-\frac{25}{4}$

vertex $(-\frac{11}{2}$ , $-\frac{25}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

******************************************************************************************

2) y = -x² - 6x - 8

a = -1 so the parabola opens DOWN

x = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3

y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1

vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

******************************************************************************************

3) y = x² - 2x + 3

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1

y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). <em>There are no x-intercepts</em>

******************************************************************************************

4) y = x² + 4x + 4

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2

y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0

vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). <em>The x-intercept is the vertex.</em>

Compared to the other four graphs, this is most likely the equation for the rain gauge!

******************************************************************************************

5) y = 3x² + 21x + 30

a = +3 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$

y = 3 $(-\frac{7}{2})^{2}$ + 21 $(-\frac{7}{2} )$ + 30 = $-\frac{27}{4}$

vertex $(-\frac{7}{2}$ , $-\frac{27}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

*******************************************************************************************

4 0

3 years ago

Read 2 more answers

How do you Find the acute Angle A when sinA=0.616?

kirill [66]

Answer:

arcsin0.616

Step-by-step explanation:

arcsino.616

4 0

3 years ago

I need help please thank you hurry

Elena-2011 [213]

Answer:

Step-by-step explanation:

To get the other angles you need to do this

110+80+x+x=360

190+2x=360

2x=360-190

2x=170

X=170÷2

X=85

5 0

3 years ago

Almonds cost $15.50 for 6 pounds. Find the unit rate. Round to the nearest cent if necessary.

Salsk061 [2.6K]

15.50 for 6 lbs......15.50/6 = $ 2.58 per lb

4 0

3 years ago