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nlexa [21]
3 years ago
5

Solve for x: |4x + 12| = 16 (5 points) x = 7, x = −7 x = 1, x = −1 x = 1, x = −7 x = −1, x = 7

Mathematics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

x =1     x = -7

Step-by-step explanation:

|4x + 12| = 16

Absolute value equations have two solutions, one positive and one negative

4x+12 = 16          4x+12 = -16

Subtract 12 from each side

4x+12-12 = 16-12          4x+12-12 = -16-12

4x =4                               4x =-28

Divide by 4

4x/4 = 4/4                       4x/4 = -28/4

x =1                                   x = -7

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What is the perimeter of a square whose diagonal is 3 square root 2? Show all work on how you got your answer
Alika [10]

Answer:

12

Step-by-step explanation:

Given: Diagonal of square= 3\sqrt{2}

To find the perimeter of square, we need to find the length of sides of square.

∴ Using the formula of diagonal to find side of square.

Formula; Diagonal= s\sqrt{2}

Where, s is side of square.

⇒ 3\sqrt{2} = s\sqrt{2}

Dividing both side by √2

⇒s= \frac{3\sqrt{2} }{\sqrt{2} }

∴s= 3

Hence, Length of side of square is 3.

Now, finding the perimeter of square.

Formula; Perimeter= 4s

⇒Perimeter= 4\times 3

∴ Perimeter= 12

Hence, Perimeter of square is 12.

6 0
4 years ago
Find the value of x
zloy xaker [14]

Answer:

66

Step-by-step explanation:

EHG is a straight line so it is 180°.

You already know the value of FHG, so you can find the value of 2x.

180° = 48° + 2x

2x = 180° - 48°

    = 132

x = 132° ÷ 2

  = 66

5 0
3 years ago
Read 2 more answers
Kaliska is jumping rope. The vertical height of the center of her rope off the ground R(t) (in cm) as a function of time t (in s
xz_007 [3.2K]

Answer:

R (t) = 60 - 60 cos (6t)

Step-by-step explanation:

Given that:

R(t) = acos (bt) + d

at t= 0

R(0) = 0

0 = acos (0) + d

a + d = 0 ----- (1)

After \dfrac{\pi}{12} seconds it reaches a height of 60 cm from the ground.

i.e

R ( \dfrac{\pi}{12}) = 60

60 = acos (\dfrac{b \pi}{12}) +d --- (2)

Recall from the question that:

At t = 0, R(0) = 0 which is the minimum

as such it is only  when a is  negative can acos (bt ) + d can get to minimum at t= 0

Similarly; 60 × 2 = maximum

R'(t) = -ab sin (bt) =0

bt = k π

here;

k  is the integer

making t the subject of the formula, we have:

t = \dfrac{k \pi}{b}

replacing the derived equation of k into R(t) = acos (bt) + d

R (\dfrac{k \pi}{b}) = d+a cos (k \pi) = \left \{ {{a+d  \ for \ k \ odd} \atop {-a+d \ for k \ even}} \right.

Since we known a < 0 (negative)

then d-a will be maximum

d-a = 60  × 2

d-a = 120 ----- (3)

Relating to equation (1) and (3)

a = -60 and d = 60

∴ R(t) = 60 - 60 cos (bt)

Similarly;

For R ( \dfrac{\pi}{12})

R ( \dfrac{\pi}{12}) = 60 -60 \ cos (\dfrac{\pi b}{12}) =60

where ;

cos (\dfrac{\pi b}{12}) =0

Then b = 6

∴

R (t) = 60 - 60 cos (6t)

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Keith_Richards [23]

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Closest you can get w/out going over.

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nirvana33 [79]
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6 0
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