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Ksivusya [100]
3 years ago
12

Solve log(base 8)(2)+log(base 8)(4x2)=1

Mathematics
1 answer:
atroni [7]3 years ago
8 0
When you have two log with same base which are adding to each other you can say: 
log(base 8)(2 * 4x^2) = 1 
so u multiplied them to each other:
log(base 8)(8x ^2) = 1 ⇒ so now you can say: 
8^1 = 8x ^2 ⇒ 8 = 8x^2 ⇒ 1 = x^2 ⇒ x =1 :)))
i hope this is helpful
have a nice day 
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40 points I'll mark you as BRAINLIEST for the person with the right answers .
Karo-lina-s [1.5K]
9. y=-1/4x^2+4x-19
group
y=(-1/4x^2+4x)-19
undistribute -1/4
y=-1/4(x^2-16x)-19
take 1/2 of -16 and squer it to get 64 then add neg and pos inside
y=-1/4(x^2-16x+64-64)-19
factorperfect square
y=-1/4((x-8)^2-64)-19
expand
y=-1/4(x-8)^2+16-19
y=-1/4(x-8)^2-3
vertex is (8,-3)

10.
group
y=(1/4x^2-3x)+18
undistribute
y=1/4(x^2-12x)+18
take 1/2 of -12 and square it and add neg and pos isndie
y=1/4(x^2-12x+36-36)+18
factor
y=1/4((x-6)^2-36)+18
expand
y=1/4(x-6)^2-9+18
y=1/4(x-6)^2+9
get to form (x-h)^2=4p(y-k)
minus 9 both sides and times 4
(x-6)^2=4(y-9)
(x-6)^2=4(1)(y-9)
so 1>0 so opens up and focus is 1 above vertex
vertex is (6,9)
so focus i (6,10)





11.
y=(-1/6x^2+7x)-80
y=(-1/6)(x^2-42x)-80
take 1/2 of linear coefient and squer it and add negative and positive inside
-42/2=-21, (-21)^2=441
y=(-1/6)(x^2-42+441-441)-80
factor perfect square the square
y=(-1/6)((x-21)^2-411)-80
expand
y=(-1/6)(x-21)^2+73.5-80
y=(-1/6)(x-21)^2-6.5
add 6.5 to both sid
y+6.5=(-1/6)(x-21)^2
times both sides by -6
-6(y+6.5)=(x-21)^2
(x-21)^2=-6(y+6.5)
(y-21)^2=4(-3/2)(y-(-6.5))
vertex is
-3/2<0 so directix is above
it is -3/2 or 1.5 units above the vertex
up is y so
-6.5+1.5=-5
the directix is y=-5




11.
in form (y-1)^2=4p(x+3)
opens left or right
(y-1)^2=4(4)(x+3)
vertex is (-3,1)
4>0 so opens right
dirextix is to left
it is 4 units to left
(-3,1)
left right is x
4 left of -3 is -4-3=7
x=-7 is da directix
3 0
3 years ago
I need help pls due tomorrow pls I just need part B done plsss
lara [203]

Part A

1 day = 1/4 hours of practice

7 days = 7/4 hours of practice (multiply both sides by 7)

1 week = 7/4 hours of practice

1 week = (4+3)/4 hours of practice

1 week = (4+3)/4 hours of practice

1 week = (4/4)+(3/4) hours of practice

1 week = 1+(3/4) hours of practice

1 week = 1 & 3/4 hours of practice

side note: 1 & 3/4 = 1.75

=======================================

Part B

Take the result from part A, and multiply it with 60

So we'll have 60 times 1&3/4 on the left side on the first line, then 60*(1+3/4) on the right side of this same line.

The rest of the lines look like this

(60*1) + (60*3/4)

60 + 60*3/4

60 + 180/4

60 + 45

105 minutes

8 0
3 years ago
This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl
Lostsunrise [7]

First,

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

and given that 90° < <em>θ </em>< 180°, meaning <em>θ</em> lies in the second quadrant, we know that cos(<em>θ</em>) < 0. (We also then know the sign of sin(<em>θ</em>), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < <em>θ</em>/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(<em>θ</em>/2) > 0 and sin(<em>θ</em>/2) > 0.

Now recall the half-angle identities,

cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / (1 + cos(<em>θ</em>))]

Notice how we don't need sin(<em>θ</em>) ?

Now, recall the Pythagorean identity:

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Dividing both sides by cos²(<em>θ</em>) gives

1 + tan²(<em>θ</em>) = 1/cos²(<em>θ</em>)

We know cos(<em>θ</em>) is negative, so solve for cos²(<em>θ</em>) and take the negative square root.

cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

Plug in tan(<em>θ</em>) = - 12/5 and solve for cos(<em>θ</em>) :

cos(<em>θ</em>) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(<em>θ</em>/2) and tan(<em>θ</em>/2) :

sin(<em>θ</em>/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(<em>θ</em>/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0
2 years ago
Helppppppppppp unicorn
TEA [102]

Answer:

angle BAC = angle DAC it is correct answer of this question ....

plz mark my answer as brainlist plzzzz vote me also so

8 0
2 years ago
Which of the following numbers is rational but not an integer?<br> -9 <br>-4.3<br> 0 <br> 3​
Bess [88]

Answer:

0

Step-by-step explanation:

The answer would be 0

7 0
3 years ago
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