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slavikrds [6]
3 years ago
9

Explain how to solve a circle for y in order to completely graph the circle.

Mathematics
1 answer:
Lelechka [254]3 years ago
5 0
The formula to graph a circle is:
(x-h)^2+(y-v)^2=r^2

Step 1: find the center if the circle.
Using this equation: x^2+y^2=r^2
Step 2: calculate the radius by solving for r
Step 3: plot the radius on the coordinate plane
Step 4: connect the dots to the graph forming a round smooth circle

Step 5: replace your (h,v) with your center points and switch the signs in your equation for the formula of graphing a circle
Step 6: calculate the radius
Step 7: plot the radius points on the coordinate plane
Step 8: connect the dots to the graph of a circle
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Please help. Urgent.
Jet001 [13]

3. ΔPQR ≅ ΔSRT

3. ASA  (Angle - Side - Angle) - we have two triangles where we know two angles and the included side are equal

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

4. PR ≅ SR

4. ΔPQR ≅ ΔSRT - the corresponding sides are congruent.

7 0
3 years ago
Find the length of the missing side. If necessary, round to the nearest tenth.
svetlana [45]

The pythagorean theorem states that the sum of the squares of two legs of a right triangle is equivalent to the hypotenuse.

So:

35^2 + b^2 = 40^2

1225 + b^2 = 1600

b^2 = 375

b = \sqrt{375}

b = 19.3649

<u>The missing side's length rounded to the nearest tenth is Option C, 19.4</u>

3 0
3 years ago
SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
pav-90 [236]

Answer:

-3f - 10

Step-by-step explanation:

-1 times 10 = -10

-1f times 3f = -3f

Sorry if this isn't right.

4 0
3 years ago
Read 2 more answers
(Free points)<br><br>Factorise x³ + 216y³ + 8z³ - 36xyz​
Nadusha1986 [10]

Answer:

[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]

Step-by-step explanation:

x³+216y³+8z³-36xyz

x³+(6y)³+(2z)³-3×6×2×xyz

As we know

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

Let a=x

b=6y

c=2z

Now.

[x+6y+2z][(x²+(6y)²+(2z)²-x×6y-6y×2z-x×2z]

[x+6y+2z][x²+(6y)²+(2z)²-6xy-12yz-2xz]

3 0
3 years ago
Read 2 more answers
find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large
katrin2010 [14]

A generic odd number can be written as

2k+1,\quad k \in \mathbb{Z}

Since there is an odd number every two numbers, three consecutive odd numbers will be

2k+1,\quad 2k+3,\quad 2k+5

Now let's make up the equations: the sum of the first two is

(2k+1)+(2k+3)

And 27 less than 3 times the largest is

3(2k+5)-27

These two must be the same, so we have

(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3

Subtracting 4k and 3 from both sides gives

1=2k \iff k=\dfrac{1}{2}

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0
3 years ago
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