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3 years ago

Please help, A sequence has the rule 4n+10, what is the 59th term?

Mathematics

1 answer:

djverab [1.8K]3 years ago

5 0

4* 59+ 10= 246

The 59th term is 246~

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What is the least common factor of 60,66

Ilia_Sergeevich [38]

Answer:3

Step-by-step explanation: 60/6=10

66/6=11

60/3=20

66/3=22 you double the numbers of the quotient which halves the factor.

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3 years ago

Which of the following is equal to 5/8÷2/7?

natta225 [31]

Answer:

5/8 * 7/2

Step-by-step explanation:

Whenever you divide fractions, you can use the 'keep change flip' method.

You keep 5/8, change the division sign to multiplication, and then flip the second fraction.

1. 5/8 / 7/2 - Keep 5/8

2. 5/8 * 7/2 - Change division sign

3. 5/8 * 2/7 - Flip the second fraction, 7/2 -> 2/7

5 0

3 years ago

Read 2 more answers

There are 6 Programmers and 8 Analysts working on a team at CIA. What is the unsimplified ratio of Programmers to Analysts?

Rzqust [24]

Answer: $\dfrac{6}{8}$

Step-by-step explanation:

Given

There are 6 Programmers and 8 Analysts working on a team at CIA.

Ratio of Programmers to analysts is

$\Rightarrow \dfrac{6}{8}$

Un-simplified ratio is $\dfrac{6}{8}$

Simplified ratio is $\dfrac{3}{4}$

6 0

3 years ago

A pet supply store sells dog biscuits and cat treats. The store has 27 boxes of dog biscuits and 21 pouches of cat treats/each b

Gre4nikov [31]

Answer:

There are more cats treats. 504 treats

Step-by-step explanation:

27 boxes of dog biscuits. Each 18 biscuits

27*18=486 biscuits

21 pouches of cat treats. Each 24 treats.

21*24=504

3 0

3 years ago

Consider the functions below. f(x, y, z) = x i − z j + y k r(t) = 4t i + 6t j − t2 k (a) evaluate the line integral c f · dr, wh

fredd [130]

With

$\vec r(t)=4t\,\vec\imath+6t\,\vec\jmath-t^2\,\vec k$

we have

$\mathrm d\vec r=(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

The vector field evaluated over this parameterization is

$\vec f(x,y,z)=\vec f(x(t),y(t),z(t))=4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k$

so the line integral is

$\displaystyle\int_{-1}^1(4t\,\vec\imath+t^2\,\vec\jmath+6t\,\vec k)\cdot(4\,\vec\imath+6\,\vec\jmath-2t\,\vec k)\,\mathrm dt$

$=\displaystyle\int_{-1}^1(16t+6t^2-12t^2)\,\mathrm dt=-4$

6 0

3 years ago