Answer:
(x + 2)² + (y + 1)² = 9
Step-by-step explanation:
Gina started with a circle with the given equation as,
(x - 2)² + (y - 0)² = 16
Since, standard equation of a circle is,
(x - a)² + (y - b)² = r²
Here (a, b) is the center of the circle and 'r' is the radius.
Therefore, from the given equation of the circle,
Center of the circle → (2, 0)
Radius of the circle → √16 = 4 units
She translated this circle 4 units left and 1 unit down.
Rule for the new center after translation will be,
(x, y) → (x - 4, y - 1)
(2, 0) → (2 - 4, 0 - 1)
→ (-2, -1)
Coordinates of the new center → (-2, -1)
Then she reduced the radius by 1 units.
Therefore, radius of the new circle = 4 - 1 = 3 units
Now we substitute these values in the standard equation of the circle to get the equation of the new circle.
(x + 2)² + (y + 1)² = 3²
(x + 2)² + (y + 1)² = 9
<span>The y-intercept of a function is given by the initial value of a function. It is the value of the function when x = 0.
Given the functions
A. Blake is tracking his savings account with an interest rate of 5% and a original deposit of $6.
The y-intercept for this function is the initial deposit which is $6.
B.
x g(x)
0 6
1 2
2 10
The y-intersept for this function is 6 which isthe value of the function g(x) when x = 0.
C. The function h of x equals 4 to the x power, plus 3
When x = 0,
Thus the y-intecept is 4.
D. j(x) = 10(2)^x
when x = 0,
Thus, the y-intercept is 10.
Thereforen the function with the highest y-intercept is
which has a y-intercept of 10.
</span>
9514 1404 393
Answer:
4a. ∠V≅∠Y
4b. TU ≅ WX
5. No; no applicable postulate
6. see below
Step-by-step explanation:
<h3>4.</h3>
a. When you use the ASA postulate, you are claiming you have shown two angles and the side between them to be congruent. Here, you're given side TV and angle T are congruent to their counterparts, sides WY and angle W. The angle at the other end of segment TV is angle V. Its counterpart is the other end of segment WY from angle W. In order to use ASA, we must show ...
∠V≅∠Y
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b. When you use the SAS postulate, you are claiming you have shown two sides and the angle between them are congruent. The angle T is between sides TV and TU. The angle congruent to that, ∠W, is between sides WY and WX. Then the missing congruence that must be shown is ...
TU ≅ WX
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<h3>5.</h3>
The marked congruences are for two sides and a non-included angle. There is no SSA postulate for proving congruence. (In fact, there are two different possible triangles that have the given dimensions. This can be seen in the fact that the given angle is opposite the shortest of the given sides.)
"No, we cannot prove they are congruent because none of the five postulates or theorems can be used."
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<h3>6.</h3>
The first statement/reason is always the list of "given" statements.
1. ∠A≅∠D, AC≅DC . . . . given
2. . . . . vertical angles are congruent
3. . . . . ASA postulate
4. . . . . CPCTC
Answer: 170
Step-by-step explanation: Like I answered before: Angle G and Angle H are consecutive angles, meaning they both add up to 180 degrees. Thus, if 10 is the value of Angle G, Angle H must be 170.
