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Snezhnost [94]
3 years ago
12

Alex went to the grocery store and bought 5 avocados. He paid $10 and received $4.50 in change. How much did each avocado cost i

f they are each the same amount?
Mathematics
1 answer:
Delvig [45]3 years ago
3 0

Answer:

$1.10

Step-by-step explanation:

We have been that Alex went to the grocery store and bought 5 avocados. He paid $10 and received $4.50 in change.

First of all, we will find amount paid for 5 avocados by subtracting $4.50 from $10.

\text{Amount paid for 5 avocados}=\$10-\$4.50

\text{Amount paid for 5 avocados}=\$5.50

Now, we will divide $5.50 by 5 to find cost of each avocado.

\text{Cost of each avocado}=\frac{\$5.50}{5}

\text{Cost of each avocado}=\$1.10

Therefore, the cost of each avocado is $1.10.

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True or false? f (x) = 2 · (1/5)^x represents an exponential function growth.
nekit [7.7K]

Hi there!

\large\boxed{\text{False.}}

f(x) = 2(1/5)ˣ

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f(x) = a(b)ˣ where:

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7 0
3 years ago
At the beginning of the day, stock XYZ opened at $6.12. At the end of the day, it closed at $6.88. What is the rate of change of
Natalija [7]

Answer:

C is the correct option.

Step-by-step explanation:

We have been given that

stock XYZ opened at $6.12 and it closed at $6.88 and we have find the rate of change of stock.

Change in the value =| Initial - final |

Change in the value = |6.12-6.88|

Change in the value = 0.76

Hence, the rate of change of stock is given by

\frac{\text{Change in value}}{\text{initial value}}\times100\\\\=\frac{0.76}{6.12}\times100\\\\=12.4\%

C is the correct option.


7 0
3 years ago
B) T is due north of C, calculate the bearing of B from C
choli [55]

Answer:

(a) 52°

(b) 322°

Step-by-step explanation:

(a) The details of the circle are;

The diameter of the circle = AOC

The center of the circle = Point O

The point the line AT cuts the circle = Point B

The point the tangent PT touches the circle = Point C

Angle ∠COB = 76°

We have that angle AOB and angle COB are supplementary angles, therefore;

∠AOB + ∠COB = 180°

∠AOB = 180° - ∠COB

∴ ∠AOB = 180° - 76° = 104°

∠AOB = 104°

OA = OB = The radius of the circle

Therefore, ΔAOB  =  An isosceles triangle

∠OAB = ∠OBA by base angles of an isosceles triangle are equal

∠AOB + ∠OAB + ∠OBA = 180° by angle summation property

∴ ∠AOB + ∠OAB + ∠OBA = ∠AOB + ∠OAB + ∠OAB = ∠AOB + 2×∠OAB = 180°

∠OAB = (180° - ∠AOB)/2

∴ ∠OAB = (180° - 104°)/2 = 38°

∠TAC = ∠OAB = 38° by reflexive property

AOC is perpendicular to tangent PT at point C, by tangent to a circle property, therefore;

∠TCA = 90° and ΔTCA = A right triangle

∠TAC + ∠ATC + ∠TCA = 180° by angle sum property

∠ATC = 180° - (∠TAC + ∠TCA)

∴ ∠ATC = 180° - (38° + 90°) = 52°

Angle ATC = 52°

(b) In ΔABC, ∠ABC = Angle subtended by the diameter = 90°

∴ ΔABC = A right triangle

∠ABC and ∠TBC are supplementary angles, therefore;

∠ABC + ∠TBC = 180°

∠TBC = 180° - ∠ABC

∴ ∠TBC = 180° - 90° = 90°

∠TCB = 180° - (∠TBC + ∠ATC)

∴ ∠TCB = 180° - (90° + 52°) = 38°

The bearing of B from C = (360° - 38°) = 322°.

7 0
2 years ago
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