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2 years ago

NEED ASAP WILL GIVE ANYTHING

Mathematics

2 answers:

Nata [24]2 years ago

8 0

Answer:

an=5n−35

Step-by-step explanation:

The question says to show the equation in (y=mx+b) form.

So, we convert -30+(n-1)5 into (y=mx+b) form.

we get rid of the parenthesis by multiplying 5 with (n-1)

n×5 = 5n

-1×5=-5

So we are left with -30+5n-5

We solve for like terms, -30-5

so we are left with 5n-35.

an=5n-35

Now we double check if this equation is valid, the question states that the 10th term is 15 and the 14th term is 35. We plug both in, replace of n

10th term is 15

an = 5(10)-35

an = 50-35

an=15 ✔ (that checks out)

14th term is 35

an = 5(14)-35

an = 70-35

an=35 ✔ (that also checks out)

THEREFORE, the equation in (y=mx+b) form would be

an=5n−35

BabaBlast [244]2 years ago

4 0

Answer:

See below

Step-by-step explanation:

Given an AP with:

a₁₀ = 15
a₁₄ = 35

General form for nth term:

aₙ = a₁ + (n - 1)d

Apply this to the given terms:

a₁ + 9d = 15
a₁ + 13d = 35

Solve the system above by elimination, subtract the equations:

13d - 9d = 35 - 15
4d = 20
d = 5

Find the first term:

a₁ + 9*5 = 15
a₁ = 15 - 45
a₁ = - 30

Now, substitute the values of the first term and common difference into general equation:

aₙ = a₁ + (n - 1)d
aₙ = - 30 + (n - 1)*5

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Alright, Since my previous questions have not been answered, here's 50p to whoever solves this

BigorU [14]

Step-by-step explanation:

We can make 2 equations as follows:

2x = y + 16 and x + 2y = 18

x + 2y = 18 is the same as x = 18 - 2y. Hence, 2(18 - 2y) = 2x = y + 16.

2(18 - 2y) = y + 16

36 - 4y = y + 16

5y = 20

y = 4.

6 0

3 years ago

An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the sam

Genrish500 [490]

Answer:

Step-by-step explanation:

Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;

Range $R = U\sqrt{\frac{H}{g} }$

U is the velocity of the arrow

H is the maximum height reached = distance below the bullseye reached by the arrow.

R is the horizontal distance covered i.e the distance of the target from the archer.

g is the acceleration due to gravity.

Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?

On substitution,

$60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}$

Squaring both sides we have;

$(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet$

The arrow will hit the target 1.84feet below the bullseye.

5 0

3 years ago

Is k=6 a solution to the equation: 1/3k = 3

Step2247 [10]

Answer: No it's not

Step-by-step explanation:

1/3 x 6/1 = 6/3

6/3 = 2

8 0

2 years ago

Read 2 more answers

Part 1. Please show work. Please assist me with these math problems.

Vitek1552 [10]

Answer: 1) 2, 3, 4, 9, 32, 279, 8896, 2481705, 22077238784

2) 2,490,924

3) Neither

Step-by-step explanation:

$S_n=S_{n-2}\cdot (S_{n-1}-1)\quad \\\\S_1=2\quad given\\S_2=3\quad given\\S_3=S_1(S_2-1)\quad =2(3-1)\quad =4\\S_4=S_2(S_3-1)\quad =3(4-1)\quad =9\\S_5=S_3(S_4-1)\quad =4(9-1)\quad =32\\S_6=S_4(S_5-1)\quad =9(32-1)\quad =279\\S_7=S_5(S_6-1)\quad =32(279-1)\quad =8,896\\S_8=S_6(S_7-1)\quad =279(8896-1)\quad =2,481,705\\S_9=S_7(S_8-1)\quad =8896(2481705-1)\quad =22,077,238,784$

$\sum^8_{k=1}S_k\\\\=S_1+S_2+S_3+S_4+S_5+S_6+S_7+S_8\\\\= 2,490,924\\$

Neither

Not arithmetic because the difference between each of the terms is not the same.

S₂ - S₁ S₃ - S₂ S₄ - S₃

3 - 2 = 1 4 - 3 = 1 9 - 4 = 5

Not geometric because the ratios between each of the terms is not the same.

$\dfrac{S_2}{S_1}=\dfrac{S_3}{S_2}\qquad \rightarrow \dfrac{3}{2}=\dfrac{4}{3}\qquad \rightarrow \text{cross multiply to get}: 9\neq 8$

8 0

3 years ago

On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove

gogolik [260]

Velocity, distance and time:

This question is solved using the following formula:

$v = \frac{d}{t}$

In which v is the velocity, d is the distance, and t is the time.

On the first day of travel, a driver was going at a speed of 40 mph.

Time $t_1$ , distance of $d_1$ , v = 40. So

$v = \frac{d}{t}$

$40 = \frac{d_1}{t_1}$

The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles

On the second day, the velocity is $v = 60$ .

On the first day, he drove 2 more hours, which means that for the second day, the time is: $t_1 - 2$

On the first day, he traveled 20 more miles, which means that for the second day, the distance is: $d_1 - 20$

Thus

$v = \frac{d}{t}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

System of equations:

Now, from the two equations, a system of equations can be built. So

$40 = \frac{d_1}{t_1}$

$60 = \frac{d_1 - 20}{t_1 - 2}$

Find the total distance traveled in the two days:

We solve the system of equation for $d_1$ , which gets the distance on the first day. The distance on the second day is $d_2 = d_1 - 20$ , and the total distance is: