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3 years ago

3x2 - 6x + 1 = 0

Mathematics

1 answer:

Tresset [83]3 years ago

4 0

Answer:

.2 and 1.8

Step-by-step explanation:

If you graph the quadratic, can find where y = 0 or where the graph intersects the x-axis.

you can also use the quadratic formula to solve this problem.

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Find the smallest square number divisible by each of these numbers 6,9,and 15

STALIN [3.7K]

Answer:

$\frac{6}{3} = 2$

$\frac{9}{3} = 3$

$\frac{15}{3} = 5$

So the smallest number is 3. As it divides 6,9 and 15.

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3 years ago

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A college student takes the same number of credits each semester. She had 8 credits when she started, and after 7 semesters, she

boyakko [2]

Answer:

12 credits per semester

Step-by-step explanation:

she has 92 in the end but she started off with 8 so your first step is to subtract 8 from 92 which gives you 84 once you get 84 you divide that by 12 you get 7 which means she got 12 credits in each of her 7 semeters

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2 years ago

Pls help meh Area of a Circle (rounded)

skelet666 [1.2K]

Answer:

28.27 or 28 for rounded

Step-by-step explanation:

Pi r^2 pi(3)^2

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3 years ago

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WILL GIVE 30 POINTS <br><br> identify the polynomial. a^2b-cd^3

madam [21]

Answer:

this is a binomial because its separated by a negative sign

Step-by-step explanation:

3 0

3 years ago

Use a half-angle identity to find the exact value

Tatiana [17]

Given:

$\cos 15^{\circ}$

To find:

The exact value of cos 15°.

Solution:

$$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}$

Using half-angle identity:

$$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}$

$$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}$

Using the trigonometric identity: $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$$=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$

Let us first solve the fraction in the numerator.

$$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}$

Using fraction rule: $\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}$

$$=\sqrt{\frac {2+\sqrt{3}}{4}}$

Apply radical rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$

Using $\sqrt{4} =2$ :

$$=\frac{\sqrt{2+\sqrt{3}}}{2}$

$$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}$

5 0

3 years ago