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3 years ago

What the tenth term an=4n-1

Mathematics

1 answer:

swat323 years ago

5 0

* Hopefully the work below helps answer your question.Mark me the brainliest:)!!

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Someone please help with this maths problem.

NemiM [27]

The answer would be 80 because you are multiplying each number by 2. You multiply 5 x 2 = 10, 10 x 2 = 20, 20 x 2 = 40, and then 40 x 2 = 80. That is how I got the answer 80!!

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3 years ago

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Genrish500 [490]

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3 years ago

Please help fast......

Tom [10]

$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\ 1.AY=BX&\text{1. Given}\\ 2.AB \cong AB&\text{2. Reflexive Property}\\ 3. AD || BC&\text{3. Property of a square}\\ 4. \angle ABE \cong \angle AXB&\text{4. Alternate Interior Angles}\\ 5. \angle BAY \cong \angle BYA&\text{5. Alternate Interior Angles}\\6. \triangle BAX \cong \triangle ABY&\text{6. Angle-Side-Angle Theorem}\\ 7. AX \cong BY&\text{7. CPCTC}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\1. AB=CF&\text{1. Given}\\2.AB+BF=A'F&\text{2. Segment Addition Postulate}\\3.CF+BF=A'F&\text{3. Substitution Property}\\4.CF+BF+BC&\text{4. Segment Addition Postulate}\\5.A'F=BC&\text{5. Transitive Property}\\6. \angle AFE = \angle DBC&\text{6. Given}\\7. EF = BD&\text{7. Given}\\8. \triangle AFE \cong \triangle CBD&\text{8. Side-Angle-Side Theorem}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\\text{1.AC bisects }\angle BAD&\text{1. Given}\\2. \angle BAC \cong \angle DAC&\text{2. Property of angle bisector}\\3.AC = AC&\text{3. Reflexive Property}&4. \angle ACB \cong \angle ACD&\text{4. Property of angle bisector}\\5. \triangle ABC \cong \triangle ADC&\text{5. Angle-Side-Angle Theorem}\\6.BC=CD&\text{6. CPCTC}\\\end{array}$

5 0

3 years ago

Just answer # 1 please

Rudiy27

Can't see the photo fully! Sorry!

4 0

3 years ago

Find the least common multiple for 6(x+1)^3(x-4)^2 and 10(x+1)^8(x-4)^5

dolphi86 [110]

Answer:

30 $(x+1)^8 (x-4)^5$

Step-by-step explanation

In order to find the Least common multiple we write each of the factor only once from each of the expression and for the common expression we take their LCM as maximum of the exponent in all expressions

as here in the question exponent of (x+1) are 3 and 8 so we take exponent 8

likewise for (x-4) we shall take maximum of 2 and 5 which is 5

so our expression for Least common multiple will be

2X3X5 X $(x+1)^8 (x-4)^5$

30 $(x+1)^8 (x-4)^5$

7 0

3 years ago