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erik [133]
2 years ago
8

Ramon earns $1,835 each month and pays $53.80 on electricity. To the nearest tenth of a percent, what percent of Ramon's earning

s are spent on electricity each month?​
Mathematics
1 answer:
Nitella [24]2 years ago
7 0

Answer:

Sorry Iam at exam now

I know the answer but IAM writing exam

You might be interested in
Between July 28 and 30 of last year over 41.2 cm​ (about 16 ​inches) of rain fell in a city. The normal total monthly rainfall i
ZanzabumX [31]

The amount more of rain that fell in those three days than normally falls in July and August combined was 31.19 cm.

<h3>How much more rain fell in those three days?</h3>

The amount of rain in those three days was 41.2 cm.

The combined normal rain July and August is:

= 6.36 + 3.65

= 10.01 cm

The difference is therefore:

= 41.2 - 10.01

= 31.19 cm

Find out more on quantity of rainfall at brainly.com/question/13045834

#SPJ1

6 0
11 months ago
The Choco soda company offers canned drinks that hold
mote1985 [20]

Answer:

1.5 inches

Step-by-step explanation:

56.52/ 8 = 7.065

7.065 / pi =2.248859346

square root of 2.248859346 = 1.5

r = 1.5

8 0
3 years ago
I need help please. Thanks!
Karolina [17]

Answer:

A

Step-by-step explanation:

We are given the function f and its derivative, given by:

f^\prime(x)=x^2-a^2=(x-a)(x+a)

Remember that f(x) is decreasing when f'(x) < 0.

And f(x) is increasing when f'(x) > 0.

Firstly, determining our zeros for f'(x), we see that:

0=(x-a)(x+a)\Rightarrow x=a, -a

Since a is a (non-zero) positive constant, -a is negative.

We can create the following number line:

<-----(-a)-----0-----(a)----->

Next, we will test values to the left of -a by using (-a - 1). So:

f^\prime(-a-1)=(-a-1-a)(-a-1+a)=(-2a-1)(-1)=2a+1

Since a is a positive constant, (2a + 1) will be positive as well.

So, since f'(x) > 0 for x < -a, f(x) increases for all x < -a.

To test values between -a and a, we can use 0. Hence:

f^\prime(0)=(0-a)(0+a)=-a^2

This will always be negative.

So, since f'(x) < 0 for -a < x < a, f(x) decreases for all -a < x < a.

Lasting, we can test all values greater than a by using (a + 1). So:

f^\prime(a+1)=(a+1-a)(a+1+a)=(1)(2a+1)=2a+1

Again, since a > 0, (2a + 1) will always be positive.

So, since f'(x) > 0 for x > a, f(x) increases for all x > a.

The answer choices ask for the domain for which f(x) is decreasing.

f(x) is decreasing for -a < x < a since f'(x) < 0 for -a < x < a.

So, the correct answer is A.

3 0
3 years ago
Can someone help me with this?
mihalych1998 [28]
Can you put a more clear pic up hard to see
4 0
3 years ago
What are the zeroes of f(x) = x^2 + 5x + 6? (4 points)
OLga [1]

Answer:

the answer is A, if you need explanation comment

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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