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Kaylis [27]
3 years ago
14

A system of equations with a linear and quadratic equation must have one solution.

Mathematics
2 answers:
Ierofanga [76]3 years ago
8 0
That is false because, this type of system can have one solution, two solutions, or no solutions. Graph both equations on the same coordinate plane. Identify the point of intersection, if any. 
Hope I Helped : )

aev [14]3 years ago
8 0
Its true  
............
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Can someone help I know it's easy. ( picture attached)
lys-0071 [83]

Answer:

B 125.6

Step-by-step explanation:

We can use the formula for the circumference to solve.

C=2\pir

C=2\pi(20)

C=(6.28)(20)

C=125.6

8 0
3 years ago
The graph of a function is shown:
lbvjy [14]
Where is the graph? not trying to be annoying just trying to answer ur question
8 0
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Trinity collected 2,400 grams of pennies in a fundraiser. Each penny has a mass of 2.5 grams. How much money did Trinity raise?
sp2606 [1]

Answer:

960

Step-by-step explanation:

6 0
3 years ago
Is this a relation or a function explain your answer
mash [69]
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4 0
3 years ago
Find an equation of the circle that satisfies the given conditions. (Give your answer in terms of x and y.) Center at the origin
Elanso [62]

Answer:

The equation of circle is x^2+y^2=65.

Step-by-step explanation:

It is given that the circle passes through the point (8,1) and center at the origin.

The distance between any point and the circle and center is called radius. it means radius of the given circle is the distance between (0,0) and (8,1).

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula the radius of circle is

r=\sqrt{\left(8-0\right)^2+\left(1-0\right)^2}=\sqrt{65}

The standard form of a circle is

(x-h)^2+(y-k)^2=r^2         .... (1)

where, (h,k) is center and r is radius.

The center of the circle is (0,0). So h=0 and k=0.

Substitute h=0, k=0 and r=\sqrt{65} in equation (1).

(x-0)^2+(y-0)^2=(\sqrt{65})^2

x^2+y^2=65

Therefore the equation of circle is x^2+y^2=65.

5 0
3 years ago
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