Through (5,1) , slope 4/5

Determine the 10th term from the end of the AP 4, 9, 14

77julia77 [94]

Answer:

<h2>The answer is 49</h2>

Step-by-step explanation:

Since the sequence above is an arithmetic sequence

For an nth term in an arithmetic sequence

A(n) = a + ( n - 1)d

where a is the first term

n is the number of terms

d is the common difference

To find the 10th term of the sequence first find the first term and the common difference

From the question

a = 4

d = 9 - 4 = 5 or 14 - 5 = 9

Since we are finding the 10th term

n = 10

Substitute the values into the above formula

That's

A(10) = 4 +(10-1)5

= 4 + 9(5)

= 4 + 45

We have the final answer as

Hope this helps you

6 0

3 years ago

write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator. 40 to 56

masha68 [24]

Answer:

5/7

Step-by-step explanation:

A common factor of 8 can be canceled from numerator and denominator.

40/56 = (5·8)/(7·8) = (5/7)·(8/8) = (5/7)·1 = 5/7

_____

Since you know your multiplication tables, you know that 40 and 56 are both multiples of 8.

__

If you don't know your multiplication tables, you can find the greatest common divisor (GCD) of the two numbers and divide each by that. The GCD can be found using Euclid's algorithm. For that, you divide the larger by the smaller and use the remainder as the new smaller number. The original smaller number is now the larger number. For these numbers, that looks like ...

56 ÷ 40 = 1 r 16

40 ÷ 16 = 2 r 8

16 ÷ 8 = 2 r 0 . . . . . the zero remainder signals that the divisor (8) is the GCD

Now, your fraction is ...

(40/8) / (56/8) = 5/7

3 0

3 years ago

Read 2 more answers

Solve the system of linear equations.

sweet-ann [11.9K]

Answer:

dependent system
x = 2 -a
y = 1 +a
z = a

Step-by-step explanation:

Let's solve this by eliminating z, then we'll go from there.

Add 6 times the second equation to the first.

(3x -3y +6z) +6(x +2y -z) = (3) +6(4)

9x +9y = 27 . . . simplify

x + y = 3 . . . . . . divide by 9 [eq4]

Add 13 times the second equation to the third.

(5x -8y +13z) +13(x +2y -z) = (2) +13(4)

18x +18y = 54

x + y = 3 . . . . . . divide by 18 [eq5]

Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:

y = 3 -x . . . . solve eq5 for y

x +2(3 -x) -z = 4 . . . . substitute into the second equation

-x +6 -z = 4

x = 2 - z . . . . . . add x-4

y = 3 -(2 -z)

y = z +1

So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...

x = 2 -a

y = 1 +a

z = a

_____

<em>Check</em>

First equation:

3(2-a) -3(a+1) +6a = 3

6 -3a -3a -3 +6a = 3 . . . true

Second equation:

(2-a) +2(a+1) -a = 4

2 -a +2a +2 -a = 4 . . . true

Third equation:

5(2-a) -8(a+1) +13a = 2

10 -5a -8a -8 +13a = 2 . . . true

Our solution checks algebraically.

6 0

2 years ago

You bike 3 miles the first day of your training, 3.3 miles the second day, 3.9 miles the third day, and 5.1 miles the fourth

docker41 [41]

Answer:

The answer to your question is 21.9 mi

Step-by-step explanation:

Data

1st day = 3 mi

2nd day = 3.3 mi

3rd day = 3.9 mi

4rd day = 5.1 mi

7fh day = ?

Process

1.- Calculate the pattern

Day 1 2 3 4 5 6 7

MIles 3 3.3 3.9 5.1 7.5 12.3 21.9

Gradient 0.3 0.6 1.2 2.4 4.8 9.6

We notice that the gradient is duplicating so the miles that I bike the seventh day are 21.9.

8 0

2 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

tensa zangetsu [6.8K]

Answer:

$y=3t^9+C$

Step-by-step explanation:

Given: $\dfrac{dy}{dt}=27t^8$

We want to obtain the general solution of the given differential equation.

$dy=27t^8$ dt\\$Take the integral of both sides\\\int dy =\int 27t^8$ dt$\\y=\dfrac{27t^{8+1}}{8+1} +C$, (where C is the constant of integration)$\\y=\dfrac{27t^{9}}{9} +C\\\\y=3t^9+C$

The general solution of the differential equation is: $y=3t^9+C$

CHECK:

$\dfrac{d}{dt} y=\dfrac{d}{dt}(3t^9+C)=\dfrac{d}{dt}(3t^9)+\dfrac{d}{dt}(C)\\\\\text{Since derivative of a constant is zero}\\\\\dfrac{dy}{dt}=27t^{9-1}\\\dfrac{dy}{dt}=27t^8$

4 0

3 years ago