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3 years ago

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Thad ran his part of the race in 11.45 seconds, Greg ran his part in 13.012 seconds and Mike ran his part in 12.837 seconds. Wha

t is the difference between how fast Mike and Thad ran their race? A. 1.377 seconds B. 1.387 seconds C. 11.692 seconds D. 24.287 seconds

1 answer:

LenaWriter [7]3 years ago

8 0

I think.its b I hope it helps

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I need help on this one asap pliz

Doss [256]

$C(5,4)(.40)^4(1-.40)^1=5(.4)^4(.6)^1=0.0768$

That's 7.68% = 7.7% (rounded)

Note: C(5, 4) is the number of combinations of 5 things taken 4 at a time. You may have seen this written as $_5C_4$

In general, $C(n,r)=\frac{n!}{r!(n-r)!}$

8 0

3 years ago

Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z

Oliga [24]

<u>Part a)</u>

Given the expression

$54m^3n\:+\:81m^4n^2$

Apply exponent rule: $a^{b+c}=a^ba^c$

$54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn$ ∵ $m^4n^2=m^3mnn$

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

$=2\cdot \:27m^3n+3\cdot \:27m^3mnn$

Factor out the common term: 27m³n

$=27m^3n\left(2+3mn\right)$

Therefore,

$54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)$

<u>Part B)</u>

Given the expression

$15x^2y^3z+25x^3y^2z+35x^2y^2z$

Apply exponent rule: $a^{b+c}=a^ba^c$

$15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z$

Rewrite as

$=3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z$

Factor out common term 5y²x²z

$=5y^2x^2z\left(3y+5x+7\right)$

Therefore,

$15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)$

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3 years ago

What value of x is in the solution set of 2x-3>11 - 5x?

leva [86]

Answer:

3. 2 or x=2

Step-by-step explanation:

I put your question into a math calculator and this was the answer. Good luck!

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3 years ago

Lily babysits on the weekends. She

Anastaziya [24]

Answer: She babysat 7 hours

Step-by-step explanation:if you take the knowledge you have; she charges 6$ per hour. and lily mad 42$ you can divide 42 by 6 to get the answer of 7

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3 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

3 years ago