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Vanyuwa [196]
3 years ago
7

HELP ASAP thank you SO much in advance (fake and spam responses will be reported) 

Mathematics
1 answer:
valina [46]3 years ago
3 0
1.B -  - original function is y = sqrt(x). If we make sqrt(x+4), we will move the original function to the left 4. If we make sqrt(x+4)+3, additionally the original function will be moved up 3.
 
2.D - original function is y = sqrt(x). If we make sqrt(x-7), we will move the original function to the right 7. If we make 5*sqrt(x-7), additionally the original function will be expanded throw the y-axis.

3.E - original function is y = x^5. If we make -x^5 (multiply x^2 by -1), we will reflect the original function over the x-axis. If we make -x^5 - 4 , we additionally will move the original function down 4.

4.C - original function is y = x^2. If we make (x-3)^2, we will move the original function to the right 3. If we make x^2 - 5 , we will move the original function down 5.

5.A - original function is y = x^2. If we multiply x^2 by 1/3, function will be compressed about the y-axis.
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If f(x) = 2x - 5, then f(4) is<br> 17<br> 5<br> 4<br> 3
vagabundo [1.1K]

Answer:

  3

Step-by-step explanation:

The notation "f(4)" means you put 4 where x is in the algebraic expression, then do the arithmetic.

  f(4) = 2·4 -5

  = 8 -5

  = 3

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3 years ago
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4. Don designed a triangular-shaped garden. Use the diagram to answer each question.
mote1985 [20]
This is a 45-45-90 triangle or a right triangle created by the sides.

Because of this being a right triangle and 45 degrees being given already for one angle you would add 90 and 45 which will give you 135. You would subtract 135 from 180 because all the angles of a triangle add up to 180°. 180-135=45. X=45°
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MArishka [77]
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3 years ago
Find the distance between P1(4,16degrees) and P2(-2,177degrees) on the polar plane.
bazaltina [42]
Polar coordinates give the distance from the origin and the angle from the positive x axis. Cartesian coordinates give the distance from the x and y axes.

You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.

Let's convert (4, 16°) to Cartesian coordinates.

Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375

sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327<span>

So (4, 16°) </span>⇒ (3.84504678375, 1.10254942327).

Let's convert (-2, 177°)  to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)

In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).

Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.

cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951

sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248

So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).

Now we must use the distance formula with our two points.
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}
d\approx\sqrt{-1.84778771^2+-1.20722134^2}
d\approx\sqrt{3.41431942+1.45738336}
d\approx\sqrt{4.87170278}
\boxed{d\approx2.20719342}

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Naddik [55]

Answer:

The answer is C. Rounding to the nearest whole number.

Step-by-step explanation:

Please mark me brainliest

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