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2 years ago

Can someone pls do these quick

Mathematics

1 answer:

balu736 [363]2 years ago

6 0

Answer:

6. x° is approximately 21.04°

7. x° is approximately 39.56°

8. x° is approximately 58.03°

9. x° is approximately 72.85°

Step-by-step explanation:

6. In the given right triangle (a triangle with the measure of one of the interior angles equal to 90°, indicated by the small square between two sides) , we have;

The hypotenuse side length = 15

The adjacent side to the given reference angle, x° = 14

By trigonometric ratio, we have;

$cos\angle X = \dfrac{Adjacent\ leg \ length}{Hypotenuse \ length}$

$\therefore cos(x^{\circ}) = \dfrac{14}{15}$

To find the value of x°, we make use of the inverse cosine function, arccos found on a scientific calculator, as follows;

x° = arccos(14/15) ≈ 21.04°

x° ≈ 21.04°

7. In the given right triangle, we have;

The length of the opposite side to the given reference angle, x° = 19

The length of the adjacent side to the given reference angle, x° = 23

By trigonometric ratios, we have;

$Tan(\angle X) = \dfrac{Opposite \, side \ length}{Adjacent\, side \ length}$

$\therefore tan(x^{\circ}) = \dfrac{19}{23}$

Therefore;

x° = arctan(19/23) ≈ 39.56°

x° ≈ 39.56°

8. In the given right triangle, the adjacent side to the reference angle, x° and the hypotenuse side are given, therefore, we have;

x° = arccos(9/17) ≈ 58.03°

x° ≈ 58.03°

9. The opposite side to the reference angle and the hypotenuse side are given

By trigonometric ratio, we have;

$sin\angle X = \dfrac{Opposite \ leg \ length}{Hypotenuse \ length}$

$\therefore sin(x^{\circ}) = \dfrac{43}{45}$

x° = arcsin(43/45) ≈ 72.85°

x° ≈ 72.85°.

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Graph each absolute value function. State the domain, range, and y-intercept.

Paraphin [41]

Answer:

i)D: $x\in R$

ii)R: $y\ge-3$

iii) Y-int:(0,-1)

Step-by-step explanation:

i) The given absolute value function is;

$f(x)=|x+2|-3$ .

The absolute value function is defined for all real values of x.

The domain is all real numbers.

ii) The range is all y-values that will make x defined.

The given function,

$f(x)=|x+2|-3$ .

has vertex at, (-2,-3) and opens upwards.

This implies that, the minimum y-value is -3.

The range is $y\ge-3$

iii) To find the y-intercept substitute x=0 in to the function.

$f(0)=|0+2|-3$ .

$f(0)=|2|-3$ .

$f(0)=2-3$ .

$f(0)=-1$ .

The y-intercept is (0,-1)

See attachment for graph.

6 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$