Given f(x) = 3. + 1, solve for when f(3) = 7.

Find the angle θ between u = <7, –2> and v = <–1, 2>.

Dmitry_Shevchenko [17]

The angle between two vectors is:

CosФ = u - v / Magnitude(u) x magnitude(v)

Magnitude of u = SQRT(7^2 + -2^2) = SQRT(49 +4) = SQRT(53)

Magnitude of v = SQRT(-1^2 +2^2) = SQRT(1 +4) = SQRT(5)

u x v = (7 x -1) + (-2 x 2) = -7 + -4 = -11

cosФ = -11 / sqrt(53) x sqrt(5)

cosФ = -11sqrt265) / 265

Ф =cos^-1(-11sqrt265) / 265)

Ф=132.51 degrees.

7 0

4 years ago

Read 2 more answers

Plzzzz help me ??!?!?!??

True [87]

Pythagorean Theorem is for right triangles

right triangles have a 90 degree angle in them

D is a right triangle

5 0

4 years ago

Read 2 more answers

Write an equation in point-slope form of the line that passes through (3, -8) and (7,-2).

sashaice [31]

Answer:

A line that passes through (3, -8) and (7,-2).

Denote equation of that line: y = ax + b

=> Slope a can be directly determined by:

a = (7 - 3)/(-2 - -8) = 4/6 = 2/3

=> y = (2/3)x + b

This line passes (3, -8), then: -8 = (2/3)*3 + b => -8 = 2 + b => b = -10

=> y =(2/3)x - 10

There are other ways to find equation of line.

Hope this helps!

:)

3 0

3 years ago

Read 2 more answers

Spencer has a wooden post that is five and three sevenths feet long. He cut two and one half feet off the post. What is the leng

Vaselesa [24]

Answer:

2 13/14 feet

Step-by-step explanation:

The remaining length is the difference between the starting length and the amount cut off:

5 3/7 - 2 1/2 = (5 -2) +(3/7 -1/2) = 3 +((3·2-7·1)/(7·2)) = 3 -1/14

= 2 13/14 . . . feet remaining

_____

<em>Additional comment</em>

Addition and subtraction of fractions can be accomplished using a formula that does not rely on finding a least common denominator. In this case, the denominators have no common factors, so their product is, in fact, the least common denominator.

$\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}$

6 0

3 years ago

Read 2 more answers

How many extraneous solutions does the equation below have?

aev [14]

Answer:

A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.

The given equation consisting of variable , m is

$\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})$

None of the two solution

$m=\frac{3}{2}\times(-1 +\sqrt{2}) and , m=\frac{3}{2}\times(-1 -\sqrt{2})$ , is extraneous.

Here, L.H.S= R.H.S

Option A: 0→ extraneous

8 0

3 years ago

Read 2 more answers