All categories

3 years ago

What are the zeros of the quadratic function f(x) = 6x2 + 12x – 7?

Mathematics

2 answers:

omeli [17]3 years ago

8 0

Answer:

-6 ± √78\6 [0.47196014..., -2.47196014... = x]

Step-by-step explanation:

The roots [zeros] are the x-values where the graph intersects the x-axis. To find the roots (zeros), replace y with 0 and solve for x.

Anika [276]3 years ago

3 0

Answer:

x = 1 + √74/6 or x = 1 - √74/6

Step-by-step explanation:

Points to remember

Zeros of a quadratic equation ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/2a

To find the zeros of given quadratic equation

It is given that,

6x² +12x - 7 = 0

Here a = 6, b = -12 and c = -7

x = [-b ± √(b² - 4ac)]/2a

= [-12 ± √(12² - 4*6*(-7))]/2*6

= [12 ± √(144 +168 )]/12

= [12 ± √312]/12

= [12 ± 2√74]/12

= 1 ± √74/6

x = 1 + √74/6 and x = 1 - √74/6

You might be interested in

Evaluate. 1,000 squared 5 =

Ksivusya [100]

1,000,000,000,000,000

5 0

2 years ago

Read 2 more answers

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

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

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

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago

Thank you in advance for help

Romashka [77]

The answer to this question is C.

6 0

3 years ago

Read 2 more answers

Please help. Its a True or False question.

faust18 [17]

Answer:

1. False

2. True

3. True

Step-by-step explanation:

Remark

The point of this question is to encourage you to make use of your calculator. Every calculator can find square roots as far as I know.

Givens

$\sqrt{70}$ = 8.367

$\sqrt{72}$ = 8.49

$\sqrt{77}$ = 8.77

One

Point A is roughly 8.15. The square root of 70 is over 8.3 So square root 70 is to the right of point A. False.

Two

Square root 72 is 8.49. B is 8.3 and C is 8.6. The statement is True

Three

sqrt(77) = 8.77. Point D is 8.75. The answer is true, but I wouldn't want to bet money on it. The difference is slightly less than 2 100s.

4 0

2 years ago

Read 2 more answers

Juan told Sylvia he got a $0.50 raise this week and his new hourly rate will be $10.25 an hour. Sylvia wants to know what Juan’s

zhenek [66]

Answer:

The correct answer is:

r + 0.50 = 10.25: Subtract .50 from both sides. The answer is $9.75.

This is because Juan got a $0.50 raise which means that his new rate will be $0.50 more than his original rate (r).

6 0

3 years ago

Read 2 more answers