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Mnenie [13.5K]
3 years ago
9

Which statement about this figure is true?

Mathematics
1 answer:
LenKa [72]3 years ago
6 0

Answer:

It has reflectional symmetry with four lines of symmetry. It has no rotational symmetry It has point symmetry. It has rotational symmetry with an angle of rotation of 90°.

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A 3 inch by 5 inch photograph is enlarged so that the longer side is 15 inches.
valentina_108 [34]

Answer:

the length of shorter side is 9 inches.

longer side is enlarged 3times

so shorter side will also be enlarged 3 times.

5 0
3 years ago
What is the ratio unknown amount (x) invested in company A to the know amount ($24) invested in company B? Write the answer as a
lara [203]

Answer:

X/24

Step-by-step explanation:

he ratio of the unknown amount (x) invested in company A to the known amount ($24) invested in company B is X/24

7 0
3 years ago
If ABC = DEC,<br> B= 44º and E= 4x<br> x= [?]
kolezko [41]

Value of x is 11

Step-by-step explanation:

  • Step 1: Find x when ΔABC ≅ ΔDEC.

When these two triangles are congruent, their corresponding angles and sides will be equal.

⇒ ∠B = ∠E

⇒ 44° = 4x

∴ x = 44/4 = 11

5 0
3 years ago
Read 2 more answers
Guys please help me to do these!!!! im tryin’ my best but it’s so hard
Eduardwww [97]

Answer:

Step-by-step explanation:

1.

∠ACB=6 x°

∠DCE=30°

6x=30°-----------------divide both sides by 6

x=30°/6= 5°

∠BCE=180°-30°=150°----------------sum of angles on a straight line

2.

∠ABD=(4+5x)°

∠CBD=(x+2)°

(4+5x)°+(x+2)°=180°-----------sum of angles on a straight line

4+5x+x+2=180°----------------collect like terms

6x+6=180°

6x=180°-6°

6x=174°------------divide by 6 both sides

x=174°÷6= 29°

∠ABD=4+5x=4+(5*29)= 4+145=149°

∠CBD= x+2=29°+2°=31°

3.

BE bisects ∠ABE=(3x+1)°

m∠DBA=(8x-14)°

1/2 (m∠DBA)° = m∠ABE

1/2(8x-14)°=(3x+1)°

(4x-7)°=(3x+1)°

4x-7=3x+1

4x-3x=7+1-------------collecting like terms

x=8°

m∠ABE=(3x+1)°=(3*8+1)=24+1=25°

m∠DBA=(8x-14)°= (8*8-14)°=(64-14)=50°

4.

∠ADE=∠CDG

50+3x-y=x+2x-16------------collect like terms

50+16-y=x+2x-3x

66-y=3x-3x

66°=y

∠ADC +∠ADB+∠BDE

90°+50°+(3x-y)°=180°

140°+(3x-y)°=180°

3x-66°=180°-140°

3x=40°+66°

3x=106°-----------divide both sides by 3

x=106°÷3= 35.33°

∠FDG=(2x-16)°= (2×35.33° - 16° )= 54.67°

∠BDE = (3x-y)°= (3×35.33°-66°)= 105.99-66=39.99°

5.

∠ABD+∠DBC=90°

(6x+4)°+32°=90°

(6x+4)°=90°-32°

6x+4=58°

6x=58°-4°

6x=54°

x=54°÷6=9°

∠ABD= (6x+4)°+32°

∠ABD= (6×9 +4)°+32°

∠ABD= (54°+4°) + 32°

∠ABD=58°+32°=90°

6.

∠AED=∠CEB

(3x+5)°=(4y-15)°------------------form equation of equality

3x+5=4y-15

5+15=4y-3x

20=4y-3x------------------------------(1)

∠AEC=∠DEB

(y+20)°=(x+15)°

y+20=x+15

20-15=x-y

5°=x-y

5+y=x----------------------------(2)

Use equation (2) in equation (1)

20=4y-3x

20=4y-3(5+y)

20=4y-15-3y

20+15=4y-3y

35°=y

Solve for x

x=5+y=5°+35°=40°

∠AED=(3x+5)°=(3×40 +5 )=120+5=125°

∠AEC= (y+20)°= 35° + 20° =55°

8 0
3 years ago
Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decrea
maw [93]

Answer:

a. dQ/dt = -kQ

b. Q = 9e^{-kt}

c. k = 0.178

d. Q = 1.063 mg

Step-by-step explanation:

a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.

Let Q be the quantity of drug left in the body.

Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then

-dQ/dt ∝ Q

-dQ/dt = kQ

dQ/dt = -kQ

This is the required differential equation.

b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.

with t = 0, Q(0) = 9 mg

dQ/dt = -kQ

separating the variables, we have

dQ/Q = -kdt

Integrating we have

∫dQ/Q = ∫-kdt

㏑Q = -kt + c

Q = e^{-kt + c}\\Q = e^{-kt}e^{c}\\Q = Ae^{-kt}               (A = e^{c})

when t = 0, Q = 9

Q = Ae^{-kt}               \\9 = Ae^{-k0}\\9 = Ae^{0}\\9 = A\\A = 9

So, Q = 9e^{-kt}

c) Use the half-life to find the constant of proportionality k.

At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours

So,

Q = 9e^{-kt}\\4.5 = 9e^{-kX3.9}\\\frac{4.5}{9} = e^{-kX3.9}\\\frac{1}{2} = e^{-3.9k}\\

taking natural logarithm of both sides, we have

ln\frac{1}{2} = ln(e^{-3.9k})\\\\-ln2 = -3.9k\\k = -ln2/-3,9 \\k = -0.693/-3.9\\k = 0.178

d) How much of the 9 mg dose is still in the body after 12 hours?

Since k = 0.178,

Q = 9e^{-0.178t}

when t = 12 hours,

Q = 9e^{-0.178t}\\Q = 9e^{-0.178X12}\\Q = 9e^{-2.136}\\Q = 9 X 0.1181\\Q = 1.063 mg

7 0
3 years ago
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