All categories

2 years ago

Solve this question and help save my life!

Mathematics

1 answer:

vlada-n [284]2 years ago

7 0

The next two term is 13 and 16 respectively.

Step-by-step explanation:

here,

a1 = -2

a2 = 1

a3 = 4

a4 = 7

a5 = 10

So, d = a2 - a1 = 1 - (-2) = 3

= a3 - a2 = 4 - 1 = 3

Here, Common difference is same everywhere and the value of d is 3

Then,

To find 6th term of this sequence

a6 = a5 + d

= 10 + 3

a6 = 13

To find 7th term of this sequence

a7 = a6 + d

= 13 + 3

a7 = 16

Thus, The next two term is 13 and 16 respectively.

-TheUnknownScientist

You might be interested in

Which expression is equivalent to 9s + 3 x 5 -2s?

marissa [1.9K]

The answer is A. When you take the 9s and the -2s, you get 7s. Then you add the 3x5 and end up with 15. Giving you the answer 7s+15

6 0

2 years ago

Read 2 more answers

There are 2 questions PLease help need it quickley will give brainiest

musickatia [10]

Answer:

This is the answer

3 0

2 years ago

A circular hedge surrounds a sculpture on a square base. The radius of the hedge is 6x. The side length of the square sculpture

andrey2020 [161]

Hi, thank you for posting your question here at Brainly.

Since the hedge is surrounding the square base, the area of the base can be determined by subtracting the area of the base from the hedge.

A = pi*r^2 - (4x)^2
A = pi*(6x)^2 - (4x)^2
A = 2x^2*(3pi - 2)

I hope I was able to answer your question. Have a good day.

6 0

3 years ago

Find the x- and y- intercepts of parabola y=5x^2-16x+10

____ [38]

Y-INTERCEPT

$y = 5x^2 - 16x + 10$

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

$\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10$

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

$y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0$

We need to use the quadratic formula

The quadratic formula helps us find what values of $x$ make the equation = 0

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\$

The x-intercepts are at:

$(\frac{8 + \sqrt{14}}{5}, 0)\\(\frac{8 - \sqrt{14}}{5}, 0)$

5 0

1 year ago

Plz plz plz help me out!!

slava [35]

Henry swims at 1.6 laps per minute, Larry swims at 1.8 laps per minute. Larry swims faster

6 0

2 years ago