Answer:no it is not
Step-by-step explanation:
i just yea
I assume you're asked to solve
4 cos²(<em>x</em>) - 7 cos(<em>x</em>) + 3 = 0
Factor the left side:
(4 cos(<em>x</em>) - 3) (cos(<em>x</em>) - 1) = 0
Then either
4 cos(<em>x</em>) - 3 = 0 <u>or</u> cos(<em>x</em>) - 1 = 0
cos(<em>x</em>) = 3/4 <u>or</u> cos(<em>x</em>) = 1
From the first case, we get
<em>x</em> = cos⁻¹(3/4) + 2<em>nπ</em> <u>or</u> <em>x</em> = -cos⁻¹(3/4) + 2<em>nπ</em>
and from the second,
<em>x</em> = <em>nπ</em>
where <em>n</em> is any integer.
Answer:
D
Step-by-step explanation:
using the rule of logarithms
log x - log y ⇔ log ( 
)
then
( 
) = a - d
Answer:
- A -- base 3, height 4
- B -- base 4, height 4
- C -- base 3, height 3
- D -- base 3, height 5
Step-by-step explanation:
A sort of straightforward way to approach this is to define base and height variables for each of the figures: ba, ha, bb, hb, bc, hc, bd, hd. Then the statements can be used to make equations in these variables.
0. hc = bc . . . . figure C is a square
1. ba = bd
2. ba = bb -1
3. hc = ba
4. hb = ha = bb
5. hc·bc = 9
6. (1/2)ha·ba +(1/2)hb·bb +hc·bc+hd·bd = 38
__
Using equations 0 and 5, we find ...
hc² = 9 ⇒ hc = 3
From equation 3, ...
ba = 3
From equation 2, ...
3 = bb -1 ⇒ bb = 4
From equation 1, ...
bd = 3
From equation 4, ...
ha = hb = 4
Filling in the values we know in equation 6, we get ...
(1/2)(4)(3) +(1/2)(4)(4) +9 +hd(3) = 38
23 +3hd = 38
hd = 15/3 = 5
The dimensions are:
- A -- base 3, height 4
- B -- base 4, height 4
- C -- base 3, height 3
- D -- base 3, height 5
