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3 years ago

F(7)+x F=2x+5x how do i solve this?? Someone please help exam tmrw

Mathematics

1 answer:

8090 [49]3 years ago

6 0

Answer:

49-7F

Step-by-step explanation:

F(7)+xF=2x+5x

F(7)+xF=7x

F(7)=7x-xF

F(7)=x(7-F)

substitute x=7 in the equation above

Therefore, 7(7-F)

=49-7F

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Find the equation of the line in standard form containing a point P(-3.3) and slope m= -4/3

bixtya [17]

Answer:

4x + 3y = -3.

Step-by-step explanation:

First substitute into the point-slope form:

y - y1 = m(x - x1):

y - 3 = (-4/3)(x - -3)

y - 3 = (-4/3)(x + 3)

y - 3 = -4/3 x - 4

y = -4/3 x - 4 + 3

y = -4/3x - 1 Multiply through be 3:

3y = -4x - 3

4x + 3y = -3 (answer).

5 0

3 years ago

I will give 5 points

ikadub [295]

Use the sum of cubes factoring rule

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

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$\sin^2 \theta - \sin \theta \cdot \cos\theta + \cos^2 \theta = 1 - \sin \theta \cdot \cos \theta\\\\(\sin^2 \theta + \cos^2\theta)- \sin \theta \cdot \cos\theta = 1 - \sin \theta \cdot \cos \theta\\\\1- \sin \theta \cdot \cos\theta = 1 - \sin \theta \cdot \cos \theta \ \ \checkmark\\\\$

Throughout the entire process, the right hand side stayed the same.

On the last step, I used the pythagorean identity.

8 0

1 year ago

Keeping the properties of exponents in mind to what power must your raise the expression 7 1/2 to get 7 as a result

exis [7]

Answer:

x=2

Step-by-step explanation:

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3 0

2 years ago

HELP WITH THESE QUESTIONS PART 2.

polet [3.4K]

Answer: (a) IV (b) positive (c) $\frac{\pi}{4}$ (d) C (e) $\sqrt{2}$

Step-by-step explanation:

a) $\frac{15\pi}{4}$ - $\frac{8\pi}{4}$ = $\frac{7\pi}{4}$ , which is located in Quadrant IV per the Unit Circle.

b) sec is $\frac{1}{cos}$ . Cos is the x-coordinate. The x-coordinate inQuadrant IV is positive.

c) the reference angle is the angle from $\frac{7\pi}{4}$ to 2π = $\frac{\pi}{4}$

d) since the angle is below the x-axis and the reference angle is $\frac{\pi}{4}$ , then the angle is equal to $-sec(\frac{\pi}{4}$ )

e) sec = $\frac{1}{cos}$ ⇒ sec $\frac{7\pi}{4}$ = $\frac{2}{\sqrt{2} }$ = $\frac{2}{\sqrt{2} }$ $(\frac{\sqrt{2}} {\sqrt{2}})$ = $\frac{2\sqrt{2} }{2}$ = $\sqrt{2}$