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docker41 [41]
2 years ago
12

PLEASE HELPPPPP I NEED TO GET THIS DONE BY TODAY . SOMEONE HELP MEE. ITS GEOGEBRA

Mathematics
1 answer:
maksim [4K]2 years ago
5 0

Answer:

It would look like this:

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Equation for "five elevenths times a number is 55"
nika2105 [10]
Let the unknown number be x.

5/11 * x = 55

Since x is being multiplied by 5/11, and you want x alone, you need to multiply both sides of the equation by the reciprocal of 5/11 which 11/5.

11/5 * 5/11 * x = 55/1 * 11/5

x = 605/5

x = 121

Answer: The number is 121.
4 0
3 years ago
Find the sum of the arithmetic sequence. 5,7,9,11,...,23
eduard

Answer:

140

Step-by-step explanation:

The arithmetic series is 5, 7, 9, 11, ........., 23.

First u have to determine the no. of terms that can be done by using

Tₙ = [a + (n - 1)d]

Tₙ-------nth term

a---------first term

n---------no.of terms in the series

d---------common difference

here a = 5,d = 2.

let it contain n terms Tₙ= [a + (n-1)d]

Substitute Tₙ, a, and d in the equation

23 = 5 + (n - 1)2

Subtract 5 from each side.

18 = (n-1)2

Divide each side by 2

(n - 1) = 9

Add 1 to each side

n = 9 + 1 = 10

The sum of the arithmetic sequence formula: Sₙ= (n/2)[2a+(n-1)d]

Substitute Sₙ, a, n and d in the equation

Sₙ= (10/2)[2(5) + (10-1)2]

Sₙ= (5)[10 + (9)2]

Sₙ= 5[10 + 18]

Sₙ= 5[28] = 140

Therefore the sum of the arithmetic sequence is 140.

6 0
2 years ago
4. Benson sells vacuum cleaners on a commission. He earns $75 for each vacuum cleaner he sells.
Wittaler [7]
He would have to sell at least 19 vacuums.
8 0
3 years ago
Read 2 more answers
Jake gave the cashier $50 to pay for 5 pairs of jeans. The cashier gave him
levacccp [35]

Answer:

8.85

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
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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