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elena55 [62]
3 years ago
8

Twenty is no greater than the sum of a number and -4.

Mathematics
1 answer:
Aleksandr [31]3 years ago
3 0

Hey there!

The statement above would be written like so...

20 < n + -4

Hope this helps you!

God bless ❤️

Happy Halloween!

xXxGolferGirlxXx

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Boxes of bouncy balls with 500 balls are sold to stores for $75.00. One store sells them to customers for 35 cents each. What is
Contact [7]

Answer:

.20

Step-by-step explanation:

75/500 = .15

35-15=20

3 0
2 years ago
If Teresa counts 9 leaves on September 1, how will the number of leaves change from may 1 to September 1
IRINA_888 [86]
Teresa counted 2 leaves on May 1st and 9 leaves on September 1st.

The difference in the number of leaves is:
9 - 2 = 7

Hence, the number of leaves has increased by 7 units.
5 0
3 years ago
An account earns simple interest. $1675 at 4.6% for 4 years a. Find the interest earned. $___ b. Find the balance of the account
Colt1911 [192]
A. First, divide 1675 by 100 to find 1% of it. 
1675 / 100 = 16.75 
Then, multiply 16.75 by 4.6 to get 4.6%. 
16.75 x 4.6 = 77.05 
77.05 is the interest for one year. 
Multiply it by 4 to get the interest for 4 years. 
77.05 x 4 = 308.20 
The interest earned is $308.20. 

B. Add the interest to the original balance to get the total. 
308.20 + 1675 = 1983.20 
The balance of the account is $1983.20 after 4 years. 

Hope this help!
7 0
3 years ago
Express each expression as a single logarithm.​
andreev551 [17]

Answer:

1. 6

2. 23

Step-by-step explanation:

1.

logx+1 ( x^2 + 2x + 1)^3

= 3logx+1  ( x^2 + 2x + 1)

= 3 logx+1 (x + 1)^2

= 6 logx+1 (x + 1)  Now the logx+1 x+1 = 1  so

The answer is 6.

2.

log11 11^23

= 23 log11  11

As in question 1 log11 11 = 1

So the answer is 23.

8 0
3 years ago
The rate at which rain accumulates in a bucket is modeled by the function r given by r(t)=10t−t^2, where r(t) is measured in mil
mars1129 [50]

Answer:

36 milliliters of rain.

Step-by-step explanation:

The rate at which rain accumluated in a bucket is given by the function:

r(t)=10t-t^2

Where r(t) is measured in milliliters per minute.

We want to find the total accumulation of rain from <em>t</em> = 0 to <em>t</em> = 3.

We can use the Net Change Theorem. So, we will integrate function <em>r</em> from <em>t</em> = 0 to <em>t</em> = 3:

\displaystyle \int_0^3r(t)\, dt

Substitute:

=\displaystyle \int_0^3 10t-t^2\, dt

Integrate:

\displaystyle =5t^2-\frac{1}{3}t^3\Big|_0^3

Evaluate:

\displaystyle =(5(3)^2-\frac{1}{3}(3)^3)-(5(0)^2-\frac{1}{3}(0)^3)=36\text{ milliliters}

36 milliliters of rain accumulated in the bucket from time <em>t</em> = 0 to <em>t</em> = 3.

4 0
3 years ago
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