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3 years ago

13

The graph of a quadratic function intercepts the x-axis in the two places and the y-axis in one place.

2 answers:

wolverine [178]3 years ago

4 0

Correct Answer:
Option 3: <span>The quadratic function has two distinct real zeros.

The function is quadratic, therefore it can have only 2 zeros. The knowledge of x-intercepts is needed to determine the zeros, y-intercepts has nothing to do with the zeros of a function. The given function has 2 unique x-intercepts, so according to the fundamental theorem of algebra, this function has 2 distinct real roots as number of distinct real roots are equal to the number of x-intercepts. Therefore, option 3 is the correct answer. </span>

hichkok12 [17]3 years ago

4 0

Answer:

3: The quadratic function has two distinct real zeros

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I have some geometric sequence questions, will give 5 points for every answer and will give Brainliest!

kherson [118]

Answer:

Step-by-step explanation:

1) since the sixth term is 3 and the fifth term 24, the common ratio would be 3/24 = 1/8

The formula for finding the nth term of a geometric sequence is

Tn = ar^(n - 1)

If t6 = 3,r = 1/8, then

3 = a × 1/8^(6 - 1) = a × (1/8)^5

a = 3/(0.125)^5 = 98304

The first term is 98304.

Second term is 98304 × 1/8 = 12288

Third term is 12288 × 1/8 = 1536

Third term is 1536 × 1/8 = 192

2) t1 = 4

t2 = - 3t(2- 1) = - 3t1 = - 3 × 4 = - 12

t3 = - 3t(3- 1) = - 3t2 = - 3 × - 12 = 36

t4 = - 3t(4- 1) = - 3t3 = - 3 × 36 = - 108

3) let the numbers be t2,t3 and t4

The sequence becomes

1/2, t2,t3, t4,8

The formula for finding the nth term of a geometric sequence is

Tn = ar^(n - 1)

8 = 1/2 × r^(5 - 1)

8 = 1/2 × r^4

16 = r^4

2^4 = r^4

r = 2

t2 = 1/2 × 2 = 1

t3 = 1 × 2 = 2

t4 = 2 × 2 = 4

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3 years ago

What is the equation of the circle whose diameter has endpoints (8,-2) and (-2,6)?

klemol [59]

Answer:

$(x-3)^{2} +(y-2)^{2}=41$

Step-by-step explanation:

step 1

Find the diameter of the circle

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$A(8,-2)\\E(-2,6)$

substitute the values

$d=\sqrt{(6+2)^{2}+(-2-8)^{2}}$

$d=\sqrt{(8)^{2}+(-10)^{2}}$

$d=\sqrt{164}\ units$

$d=2\sqrt{41}\ units$

step 2

Find the center of the circle

The center is the midpoint of the diameter

The center is equal to

$C=(\frac{8-2}{2},\frac{-2+6}{2})$

$C=(3,2)$

step 3

Find the equation of the circle

The equation of the circle in center radius form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

we have

(h,k)=(3,2)

$r=2\sqrt{41}/2=\sqrt{41}\ units$ ---> the radius is half the diameter

substitute

$(x-3)^{2} +(y-2)^{2}=(\sqrt{41})^{2}$

$(x-3)^{2} +(y-2)^{2}=41$

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4 years ago

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bixtya [17]

Answer:

undefined

Step-by-step explanation:

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3 years ago

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Evaluating expressions k3+j

mrs_skeptik [129]

The answer is 3k+j since 3 cannot be multiplied after k.

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What’s the answer to this question

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The answer is for question 12, x = -9.3333333

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