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KengaRu [80]
2 years ago
12

Three consecutive integers have a sum of 72 . Find the integers.

Mathematics
1 answer:
ANTONII [103]2 years ago
6 0

Answer:

23, 24 and 25

Step-by-step explanation:

These are consecutive numbers that add up to 72.

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Laura walked 6 1/2 laps of a circular path in 2 hours. At that rate, how many laps of the
NARA [144]

Answer:

3.25 in one hour

Step-by-step explanation:

6.5 ÷ 2 = 3.25

6 0
3 years ago
ABC is a tangent to the circle below. O is the centre of the circle. Work out the size of angle 0. Give reasoning to justify you
Delvig [45]

Answer:

Θ = 46°

Step-by-step explanation:

the angle between a tangent and a radius at the point of contact is 90° , so

∠ ABO = 90°

since OB = OD ( radii of circle ) then Δ BOD is isosceles and

∠ OBD = ∠ ODB = 22°

the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ AOB is an exterior angle of the triangle , then

∠ AOB = 22° + 22° = 44°

the sum of the 3 angles in Δ AOB = 180° , then

Θ + 44° + 90° = 180°

Θ + 134° = 180° ( subtract 134° from both sides )

Θ = 46°

6 0
1 year ago
Solve for Y to get the equation into slope-intercept form and then state the slope and state the slope and state the y-intercept
valentinak56 [21]
-8-4y=-5x
-4y=-5x+8
y=5/4x+2

slope= 5/4
y-intercept=(0,2)
4 0
3 years ago
Read 2 more answers
If g (x) = 1/x then [g (x+h) - g (x)] /h
lys-0071 [83]

Answer:

\dfrac{-1}{x(x+h)}, h\ne 0

Step-by-step explanation:

If g(x) = \dfrac{1}{x}, then g(x+h) = \dfrac{1}{x+h}. It follows that

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

  = \dfrac{-1}{x(x+h)}, h\ne 0

5 0
3 years ago
Eight 5 bills are worth 40
dmitriy555 [2]

Answer:

what is the question

Step-by-step explanation:

5 0
2 years ago
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