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KonstantinChe [14]
2 years ago
7

A line intersects the points

Mathematics
1 answer:
Nimfa-mama [501]2 years ago
5 0
Y= -x+1 because the slope is negative
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The equation |p| = 2 represents the total number of points that can be earned or lost during one turn ofa game. Which
amid [387]
-2 or 2 is the answer. The absolute value of 2 is 2 and the absolute value of -2 is 2 because absolute value can not be negative.
5 0
3 years ago
Given the graph of f'(x) shown below, find the intervals on which the function f(x) is increasing.
AURORKA [14]

Answer:

intervals (-3,-1) and (0,+infinity)

Step-by-step explanation:

if f'(x)>0 then f is increasing

3 0
2 years ago
What is the equivalent decimal form for<br> 2/40?
Lelechka [254]

Answer:

decimal form: 0.05

Step-by-step explanation:

To find decimals divide them, so 2 divided by 40 is 0.05.

I hope this helped and if it did I would appreciate it if you marked me Brainliest. thank you and have a nice day!

4 0
3 years ago
Read 2 more answers
Help please I don’t understand
MaRussiya [10]

set them equal to their sum. which in this case is 141 then solve for x and plug it into your original equation to get your answer.

3 0
3 years ago
Help with 5b please . thank you.​
Allushta [10]

Answer:

See explanation

Step-by-step explanation:

We are given f(x)=ln(1+x)-x+(1/2)x^2.

We are first ask to differentiate this.

We will need chain rule for first term and power rule for all three terms.

f'(x)=(1+x)'/(1+x)-(1)+(1/2)×2x

f'(x)=(0+1)/(1+x)-(1)+x

f'(x)=1/(1+x)-(1)+x

We are then ask to prove if x is positive then f is positive.

I'm thinking they want us to use the derivative part in our answer.

Let's look at the critical numbers.

f' is undefined at x=-1 and it also makes f undefined.

Let's see if we can find when expression is 0.

1/(1+x)-(1)+x=0

Find common denominator:

1/(1+x)-(1+x)/(1+x)+x(1+x)/(1+x)=0

(1-1-x+x+x^2)/(1+x)=0

A fraction can only be zero when it's numerator is.

Simplify numerator equal 0:

x^2=0

This happens at x=0.

This means the expression,f, is increasing or decreasing after x=0. Let's found out what's happening there. f'(1)=1/(1+1)-(1)+1=1/2 which means after x=0, f is increasing since f'>0 after x=0.

So we should see increasing values of f when we up the value for x after 0.

Plugging in 0 gives: f(0)=ln(1+0)-0+(1/2)0^2=0.

So any value f, after this x=0, should be higher than 0 since f(0)=0 and f' told us f in increasing after x equals 0.

8 0
2 years ago
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