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sdas [7]
2 years ago
14

What is the value of f(0)

Mathematics
1 answer:
DIA [1.3K]2 years ago
7 0
The answer is zero because any number multiplied by 0 is 0 (the multiplicative property of zero)

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Could 10.3cm 4.4cm and 8.3 cm be the side lengths of a triangle yes or no
evablogger [386]

Answer:

No

Step-by-step explanation:

As you probably know, triangles have 3 sides. The longest side is called the hypotenuse. In this case, the hypotenuse is 10.3. Now, you might or might not know the pythagorean theorem, which states that the <em>a² + b² =c ². </em>

In this case, we can say that 4.4 is <em>a </em>and 8.3 is <em>b. </em>Now if we square 4.4 and add it to 8.3 squared, we get 88.25. However, if you square 10.3, you get 106.09. Thus, the values cannot be this way. So, your answer is no.

3 0
2 years ago
Which equation does the graph represent (see picture)
egoroff_w [7]
Y=Mx +b

B is 0 because that’s where the graph intercepts the y axis.

M is found by picking any two points and finding the rise and run because m is equal rise/run

Points of choice: (-2,4) and (2,-4)

=2−1/2−1 = -8/4

m is equal -2

Y= -2X

Look at the picture below to differentiate between positive and negative slopes.

8 0
3 years ago
What are the x and y coordinates for 6y + 4x = 12 and -6x + y =4
Licemer1 [7]
6y+y=6 
4x+-6x=-2
is ur awnser

5 0
3 years ago
Answer the question
dmitriy555 [2]
Pi is canceled out
so 3/4 x 180 = 135
hope it helps
3 0
3 years ago
A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set
lara [203]

A circle with center (3,0) and radius 5 has equation

(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x  = 16

If we substitute y=2x+k in this equation, we have

x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0

The expression simplifies to

-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89

The solutions to the associated equation are

k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}

So, the parabola is negative between the two solutions:

-6-5\sqrt{5}

8 0
3 years ago
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