#21 please I need help

Ricky has a 26-foot ladder. He must have the bottom of the ladder 10 feet from the wall. Will he be able to reach a roof that is

matrenka [14]

Check the picture below.

the ladder is 26ft, the roof is 24ft above, so the ladderis 2ft longer than needed, if you move the ladder 10 feet away, will it give 24 or more? well, you know already.

6 0

3 years ago

Please help me fill in the blanks :) marking most Brainly

Gala2k [10]

Answer:

For the part that contains organic matter your answer is, 35 negative result

For the part that does not contain organic matter your answer is, 60 positive result

Step-by-step explanation:

If we pay attention to the part that contains organic matter the total amount 700 and 665 is positive, so subtract them 700-665 to get 35 negative result

For the part that does not contain organic matter the total amount is 300, and 240 of it is negative, so subtract 240 from 300 to get 60 positive result

Your Welcome, Hope This Helps, And Have an Awesome Day!!!

6 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

2 years ago

Fill in the blank.

Murljashka [212]

Cycle of rotation hope this helps

6 0

3 years ago

Read 2 more answers

Math help guys how wouls i work this out

Andrei [34K]

8 > 7 + $\frac{x}{6}$ Subtract 7 from both sides

1 > $\frac{x}{6}$ Multiply both sides by 6

6 > x Flip it around so it's easier to read

x < 6

You can graph your answer by drawing an open circle at the 6 and coloring the line to the left. The circle should be open, because x is <em>less than 6</em>, not less than or equal to. You would color to the left to show that x can be anything less than 6.

4 0

3 years ago