Be sure to include the "=" sign: <span>f(x) = 2^x -7

To find the x-intercepts, set f(x) = 0 and solve for x: 2^x - 7 = 0,
or
2^x = 7

Take the common log of both sides: x log 2 = log 7
log 7
Solve for x: x = --------- = 2.81 (approx)
log 2

</span>

7 0

3 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

3 years ago

When f(x)=5x^2-2x+5, evaluate f(-3)

patriot [66]

Answer:

f(-3)=56

Step-by-step explanation:

When evaluating a specific input for a function, you just plug that input in for x.

So in this case, f(-3)=5(-3)^2-2(-3)+5

f(-3)=5(-3)^2-2(-3)+5

f(-3)=5(9)+6+5

f(-3)=45+6+5

f(-3)=56

8 0

3 years ago