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kirill [66]
3 years ago
12

44 children on a trip. each adult can supervise up to 10 children. maximum group size is 12 people. 2 childrebn each have their

own carer. These adult carers cannot supervise any other children. The school wants the smallest number of groups. organise the adults and the children into groups for the visit.
Mathematics
1 answer:
max2010maxim [7]3 years ago
5 0
Group 1: one Adult with 9 children group size =10
Group 2: one Adult with 9 children Group size = 10
Group 3: one Adult with 9 children Group size =10
Group 4: one Adult with 9 children Group size = 10
Group 5: one Adult with 6 children plus 2 children with their carer  Group size = 11

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Determining and proving congruence
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Answer:

Two triangles are congruent if they have: exactly the same three sides and. exactly the same three angles

There are five ways to find if two triangles are congruent: SSS, SAS, ASA, AAS and HL.

SSS (side, side, side) ...

SAS (side, angle, side) ...

ASA (angle, side, angle) ...

AAS (angle, angle, side) ...

HL (hypotenuse, leg)

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Salma Serra purchased a truck. last year she drove the truck 12,340 miles. the fixes costs totaled $1,428 , while variable cost
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A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form
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\vec{v}= \text{ or } \approx

Step-by-step explanation:

Component form of a vector is given by \vec{v}=, where i represents change in x-value and j represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector \vec{v}=, the magnitude is ||v||=\sqrt{i^2+j^2.

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being i, one leg being j, and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}

To find the other leg, i, we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}

Verify that (9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark

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