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notsponge [240]

3 years ago

6

write a two step problem that you can solve using the strategy draw a diagram.explain how can use the strategy to find the solut

ion

1 answer:

nevsk [136]3 years ago

3 0

F(x)=2x+1
you can use a table of values for a diagram and graph the line

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This is for my Precalculus teacher and I am kinda confuse, help please!!

Andrews [41]

Answer:

They are different

6 0

3 years ago

Which of the following is an equation of the line through (2,3) and (-1,-12)

ahrayia [7]

Answer:

y = 5x - 7

Step-by-step explanation:

Using the equation of a line with two points

y_2 - y_1 / x_2 - x_1 = y - y_1 / x - x_1

Giving two points

( -4 , 10)(16 , -2)

x_1 = -4

y_1 = 10

x_2 = 16

y_2 = -2

Using the above formula

y_2 - y_1 / x_2 - x_1 = y - y_1 / x - x_1

-12 - 3 / -1 - 2 = y - 3 / x - 2

-15/-3 = y - 3 /x - 2

Cross multiply

-15(x - 2) = -3(y - 3)

-15x + 30 = -3y + 9

-15x + 30 + 3y - 9 = 0

-15x +30 - 9 + 3y = 0

-15x + 21 + 3y = 0

3y = 15x - 21 ( following y = mx + C

Dividing through by 3 to find y

3y/3 = 15x -21 / 3

y = 15x - 21 / 3

We can still separate

y = 15x / 3 - 21/3

y = 5x - 7

Therefore, the equation of the line is

y = 5x - 7

6 0

3 years ago

PLS HELP WILL MARK U BRAINLIEST!! NO FAKE ANSWERS

erica [24]

Answer:

$\huge\boxed{\sf 3(n + 5) \leq 100}$

Step-by-step explanation:

The sum of a number n and 5 => n + 5

Thrice the sum of a number n and 5 => 3(n + 5)
At most means "less than or equal to", the sign of which is "≤"

<u>So, the inequality will be:</u>

3(n + 5) ≤ 100

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

6 0

2 years ago

Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif

xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0

3 years ago

In the sequence

egoroff_w [7]

Answer: 8

Step-by-step explanation:

Sequence: 1,2,2,4,8,32,256

After the first digit, the repeated pattern for the last digit is 2,2,4,8,2,6

So, the 35 term's final digit will be 8

8 0

2 years ago