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dimulka [17.4K]
3 years ago
14

2, 9, 16, 23, Number pattern expression equations

Mathematics
2 answers:
algol [13]3 years ago
7 0

Answer:

see below

Step-by-step explanation:

2+7 = 9

9+7 = 16

16+7 = 23

We are adding 7 each time

an = a1+ 7(n-1)  where an is the nth term  a1 is the first term and n is the term number

an = 2 + 7n - 7

an = 7n -5

or

an+1 = an +7

miskamm [114]3 years ago
4 0

Answer:

Below

Step-by-step explanation:

The sequence first terms are:

● 2, 9, 16, 23

Notice that:

● 23 - 16 = 7

● 16 - 9 = 7

● 9 - 2 = 7

Each time we add 7 to get a new term

● 2 + 7 = 9

To get the next term take the last one and add 7 to it.

● 23 + 7 = 30

The next term is 30

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What’s 3/4 x 22/7 x 2
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Answer:

4\frac{5}{7}

Step-by-step explanation:

we just multiply numerator by numerator and denominator by denominator

3/4x22/7

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During the 2005 football​ season, Team Tackle beat Team Sack by 6 points. If their combined scores totaled 60​, find the individ
Travka [436]

2x+6=60

2x=60-6

2x=54

x=54/2

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27+27+6=60

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2 years ago
Pls help me solve this
melamori03 [73]

Answer:

See below.

Step-by-step explanation:

1.) False - It contains 6 sig figs, not 2.

2.) False - It goes up to 2 decimal places, not 3.

3.) True

4.) False - It goes up to 1 sig fig, not 4.

8 0
3 years ago
If D is midpoint of side BC of triangle ABC. P and Q are points lying respectively on side AB and AC such that DP is parallel to
LUCKY_DIMON [66]

See proof below

Step-by-step explanation:

Assume triangle ABC to have vertices at;

A(2,-1), B(2,-7) and C(6,-7)

D is midpoint of BC, thus D is at (4,-7)

The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA

Plot the points on a graph tool and join the points to view the sketch.

To prove area of triangle CPQ is 1/4 area of ABC will be;

Find area ABC and CPQ then compare the areas.

Apply the distance formula to find the length of sides of the triangles then find the areas.

The distance formula is;

d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}

Length of side AB from the sketch is;

AB=\sqrt{(2-2)^2+(-7--1)^2} =\sqrt{-6^2} =\sqrt{36} =6units

Length of side BC will be;

BC=\sqrt{(-7--7)^2+(6-2)^2} =\sqrt{4^2} =\sqrt{16} =4units

Thus area of triangle ABC will be;

1/2 *base length*height ------because it is a right-triangle

1/2*4*6=12 square units

Find the lengths of all sides of triangle CPQ

Length of side PQ is half that of side BC thus PQ=2 units

Length of side PC is;

PC=\sqrt{(6-2)^2+(-7--4)^2} =\sqrt{4^2+-3^2} =\sqrt{16+9} =\sqrt{25} =5units

Length of side QC will be;

QC=\sqrt{(6-4)^2+(-7--4)^2} =\sqrt{2^2+-3^3} =\sqrt{4+9} =\sqrt{13}

QC= √13 = 3.6 units

Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;

A=√s(s-a)(s-b)(s-c)  where;

A=area of the triangle

s= half the perimeter of the triangle

a=side PQ = 2 units

b=side PC = 5 units

c= side QC = 3.6 units

Finding the perimeter of triangle CPQ will be;

P=sum of all sides

P=2+5+3.6 =10.6 units

s=10.6/2 = 5.3

Area of the triangle CPQ will be;

A=\sqrt{5.3(5.3-2)(5.3-5)(5.3-3.6)} \\A=\sqrt{5.3(3.3)(0.3)(1.7)} \\A=\sqrt{8.9} =2.98

A=3.0 (1 decimal place)

Compare the areas;

Area of triangle ABC=12 square units

Area of triangle CPQ = 3 square units

Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4

Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3

Learn More

Area of a triangle ;brainly.com/question/14869984

The Heron's formula : brainly.com/question/10713495

Keywords: midpoint, triangle, sides, parallel, prove , area, equal

#LearnwithBrainly

6 0
2 years ago
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