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jeka57 [31]
3 years ago
6

Is this the right Answer?

Mathematics
1 answer:
AfilCa [17]3 years ago
6 0
Check out the attached image. 

Figure 1 moves to figure 2 after the translation rule (x,y) ---> (x+1, y+2)

Figure 2 moves to figure 3 after the rotation 90 degrees clockwise around the origin

Figure 3 moves to figure 4 after the translation rule (x,y) ---> (x+2, y-3)

Figure 4 is in quadrant IV. The size does not change

-----------------------------------------------

Answer: Choice B) Quadrant IV; no

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Confused from where I left off PLEASE help
Leviafan [203]

Not sure the right equation so i did both....

#1 - √12m√15m=1m√2

Simplifies to:

13.416408m2=1.414214m

Let's solve your equation step-by-step.

13.416408m2=1.414214m

Step 1: Subtract 1.414214m from both sides.

13.416408m2−1.414214m=1.414214m−1.414214m

13.416408m2−1.414214m=0

Step 2: Factor left side of equation.

m(13.416408m−1.414214)=0

Step 3: Set factors equal to 0.

m=0 or 13.416408m−1.414214=0

m=0 or m=0.105409

Answer:  m=0 or m=0.105409


#2 - √12m√15m

Simplifies to:

13.416408m2

=13.416408*m2

=13.416408*(m*m)

=13.416408m2


Can I get a brainliest Please & Thank you...

4 0
3 years ago
Read 2 more answers
Simplify the expression where possible.<br><br> (-5 x 2 ) 3 help please!
MAXImum [283]
The answer to (-5 × 2) 3 = -30.
8 0
3 years ago
Beth left school driving toward the lake one hour before tim. tim drove in the opposite direction going 6 mph slower than beth f
m_a_m_a [10]

Beth's speed was 60 mph.

<h3>What is speed?</h3>
  • The speed (commonly referred to as v) of an object in everyday use and kinematics is the magnitude of the change in its position over time or the magnitude of the change in its position per unit of time; it is thus a scalar quantity.
  • The distance traveled by an object in a time interval is divided by the duration of the interval; the instantaneous speed is the limit of the average speed as the duration of the time interval approaches zero.
  • Velocity is not the same as speed.

To find what was Beth's speed:

Beth's data:

  • time = 2 hrs; rate = r mph; distance = r×t = 2r miles

Tim's data:

  • time = 1 hr; rate = r-6 mph; distance = r-6 miles

Equation:

  • distance + distance = 174 miles

So,

  • 2r + r - 6 = 174
  • 3r - 6 = 174
  • 3r = 180
  • r = 60 mph (Beth's rate)
  • r-6 = 54 mph (Tims's rate)

Therefore, Beth's speed was 60 mph.

Know more about speed here:

brainly.com/question/4931057

#SPJ4

3 0
1 year ago
Median of 5,4,6,3,4?
taurus [48]

Answer:

4

Step-by-step explanation:

3, 4, 4, 5, 6

7 0
3 years ago
Read 2 more answers
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
3 years ago
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