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Luda [366]
3 years ago
8

ASAP WILL MARK BRAINLYEST answer these two please!

Mathematics
1 answer:
Trava [24]3 years ago
3 0

19.\\y=3x^2\\a=3,\ b=0,\ c=0\\\\\dfrac{-b}{2a}=\dfrac{-0}{2(3)=0\to x=0\\\\3x^2=0\ \ \ \ |:3\\\\x^2=0\to x=0

20.\\y=x^2-x-2\\a=1,\ b=-1,\ c=-2\\\\\dfrac{-b}{2a}=\dfrac{-(-1)}{2(1)}=\dfrac{1}{2}\to x=\dfrac{1}{2}\\\\x^2-x-2=0\\\\x^2-2x+x-2=0\\\\x(x-2)+1(x-2)=0\\\\(x-2)(x+1)=0\iff x-2=0\ \vee\ x+1=0\\\\x=2\ \vee\ x=-1

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46h+56h=600 please solve
valentinak56 [21]

Answer:

5.88235294118

Step-by-step explanation:

Step 1:

46h + 56h = 600       Equation

Step 2:

102h = 600

Step 3:

h = 600 ÷ 102

Answer:

h = 5.88235294118

Hope This Helps :)

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3 years ago
QR is the radius of R and PQ is the tangent to R. Find the value of x.
steposvetlana [31]

Answer:

x = 39

Step-by-step explanation:

We know that PRQ is a right triangle, so we use the pythagorean theorem to solve the problem. The theorem states that a^2 + b^2 = c^2. Let's assume that the length of RP is c, the length of PQ is a and x is b. We plug the lengths into the equation and solve it:

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