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3 years ago

11

Slope of (9,2) and (6,-2)

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

m= 4/3

Step-by-step explanation:

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Find the values of which satisfy the equation. 2 cos (2B+30°) = -√3 in the domain OP ≤B≤360°

PSYCHO15rus [73]

Answer:

60 and 90

Step-by-step explanation:

$2 \cos(2 \beta + 30) = - \sqrt{3}$

$\cos(2 \beta + 30) = - \frac{ \sqrt{3} }{2}$

$2 \beta + 30 = \cos {}^{ - 1} ( \frac{ - \sqrt{3} }{2} )$

First solution.

$2 \beta + 30 = 150$

$2 \beta = 120$

$\beta = 60$

Second solution

$2 \beta + 30 = 210$

$2 \beta = 180$

$\beta = 90$

5 0

2 years ago

0.625 into a fraction

n200080 [17]

Answer:

5/8

You put 625 over 1000 because it is in the thousandths place and then simplify

5 0

4 years ago

Read 2 more answers

Given m = 2 and the point (-1, 7), which of the following is the point-slope form of the equation?

ANEK [815]

The point-slope form:

$y-y_1=m(x-x_1)$

We have the point (-1, 7) and the slope m = 2. Substitute:

$y-7=2(x-(-1))\\\\\boxed{y-7=2(x+1)}\to\boxed{b.}$

3 0

4 years ago

Determine the intercepts of the line

aleksklad [387]

Answer:

y-intercept: (0,5); x-intercept: (5/2, 0)

Step-by-step explanation:

The equation of the line is given: -4x + 7 = 2y - 3.

To find the x-intercept, set y = 0 and solve for x:

-4x + 7 = 0 - 3. This becomes -4x = -10, or x = 5/2. The x-intercept is (5/2, 0).

To find the y-inercept, set x = 0 and solve for y:

0 + 7 = 2y - 3, or 2y = 10, or y = 5. Then the y-intercept is (0, 5).

6 0

3 years ago

Please help, Brainliest if correct!!!

Lubov Fominskaja [6]

See the attached pic below

In the triangle formed by the statue and the boat, we have

$\tan23^\circ=\dfrac x{110+y}$

and

$\tan38^\circ50'=\dfrac xy$

We can solve either equation for $y$ , then substitute that into the other equation. First, let's abbreviate $t_1=\tan23^\circ$ and $t_2=\tan38^\circ50'$ . Then

$y=\dfrac x{t_2}$

$\implies t_1=\dfrac x{110+\frac x{t_2}}$

$\implies\dfrac{t_1}{t_2}=\dfrac x{110t_2+x}$

$\implies(110t_2+x)\dfrac{t_1}{t_2}=x$

$\implies110t_1=\left(1-\dfrac{t_1}{t_2}\right)x$

$\implies x=\dfrac{110t_1}{1-\frac{t_1}{t_2}}=\dfrac{110t_1t_2}{t_2-t_1}$

$\implies\boxed{x\approx99}$

7 0

4 years ago