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2 years ago

someone help!! what is the rule for the transformation formed by a translation 6 units to the left and 3 units down

Mathematics

2 answers:

ki77a [65]2 years ago

8 0

Answer: a

Because in translation when you go down the y axis is negative and when you move to the left the x axis is negative

mezya [45]2 years ago

4 0

Answer:

Step-by-step explanation:

When doing translations, moving down is represented by a negative and moving to the left is negative whereas moving up is positive and moving to the right is also positive.

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A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

2 years ago

PQR is a right angled triangle calculate the size of angle marked x

romanna [79]

Answer:

$\displaystyle x \approx 37.4^\circ$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Trigonometry

[Right Triangles Only] SOHCAHTOA
[Right Triangles Only] cosθ = adjacent over hypotenuse

Step-by-step explanation:

Step 1: Define

Angle θ = x

Adjacent Leg = 5.8

Hypotenuse = 7.3

Step 2: Solve for x

Substitute in variables [Cosine]: $\displaystyle cosx^\circ = \frac{5.8}{7.3}$
[Fraction] Divide: $\displaystyle cosx^\circ = 0.794521$
[Equality Property] Trig inverse: $\displaystyle x^\circ = cos^{-1}(0.794521)$
Evaluate trig inverse: $\displaystyle x = 37.39^\circ$
Round: $\displaystyle x \approx 37.4^\circ$

8 0

2 years ago

Solve for x. 6x2−2=13

makkiz [27]

Answer:

x = ±sqrt(5/2)

Step-by-step explanation:

6x^2−2=13

Add 2 to each side

6x^2−2+2=13+2

6x^2 = 15

Divide each side by 6

6/6x^2 = 15/6

x^2 = 5/2

Take the square root of each side

sqrt(x^2) = ±sqrt(5/2)

x = ±sqrt(5/2)

5 0

2 years ago

Read 2 more answers

Use limits to find the area of the region bounded by the graph f(x)=4-2x^3 , the x-axis , and the vertical lines x=0 and x=1

gtnhenbr [62]

Answer:

$\frac{7}{2}$ square units.

Step-by-step explanation:

We have to use limits to find the area of the region bounded by the graph $f(x) = 4 - 2x^{3}$ , the x-axis, and the vertical lines x=0 and x=1.