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3 years ago

14

20 hours put into degrees

1 answer:

dexar [7]3 years ago

5 0

Answer:

300°

Step-by-step explanation:

If 24 hours is 360°, then ...

(20 hours)/(24 hours) × 360° = 300°

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Plsss help it’s due today in an hour!!! (test grade)

Rudik [331]

Answer:

the area is 50.24 $cm^{2}$ .

Step-by-step explanation:

The formula to find area of a circle is A=π $r^{2}$ . A is area, and r is for radius. Since we have the diameter here, we just need to divide by 2 to find the radius. 8÷2=4, so there, the radius is 4. Now we need to square 4, and do 4x4, which equals 16. Now to multiply 16 by π, or 3.14. 16 x 3.14 = 50.24. The final answer is 50.24 $cm^{2}$ . Hope this helped! :]

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3 years ago

An angle measures 19.4° more than the measure of its complementary angle. What is the measure of each angle?

8090 [49]

Answer:

37.8+52.2=90

Step-by-step explanation: x + (x + 14.4) = 90

2x + 14.4 = 90

2x = 75.6

x = 37.8 (one angle)

And (x + 14.4) = 52.2 (other angle).

7 0

3 years ago

How do I find slope intercept

Nastasia [14]

Answer:

This can be done by calculating the slope between two known points of the line using the slope formula.

This can be done by substituting the slope and the coordinates of a point (x, y) on the line in the slope-intercept formula and then solve for b.

4 0

3 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago

What is the solution ? Please help with this !!

FromTheMoon [43]

Answer:

the answer is c

Step-by-step explanation:

see picture

7 0

3 years ago