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Rus_ich [418]
3 years ago
11

Five less than thw product of 3 and a number is 8

Mathematics
1 answer:
vredina [299]3 years ago
5 0
3n-5=8. For sure. Hope it helps
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What function g describes the graph of f after the given transformations?
Alex17521 [72]

After performing the transformation on f(x) = |x|+7; reflected across the x-axis and translated 4 units up. We get the function g(x) = -(|x| + 7) + 4

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

We have a function:

f(x) = |x| + 7

First we reflect the function f(x) across the x-axis

Multiply the function by -1

F(x) = -(|x| + 7)

To translated 4 units up add 4 to the function.

g(x) = -(|x| + 7) + 4

Thus, after performing the transformation on f(x) = |x|+7; reflected across the x-axis and translated 4 units up. We get the function g(x) = -(|x| + 7) + 4

Learn more about the function here:

brainly.com/question/5245372

#SPJ1

8 0
2 years ago
A factor tree for 61 that is not nothing
ddd [48]
False because the factor tree for 61 is 1 and 61

5 0
3 years ago
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

r = 0.0726

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
3 years ago
725% work in an organisation 52% of employees or post graduate and the remaining are not post graduate find the number of employ
Liono4ka [1.6K]

Answer: 348

Step-by-step explanation:

Given : Total employees work = 725

Proportion of employees are post graduate = 52%

Proportion of employees are not post graduate = 100%-52% = 48%

The number of employees were not post graduate  = 48% of 725

= 0.48 x  725

= 348

Hence, the number of employees were not post graduate  = 348

3 0
3 years ago
The first two terms of an arithmetic sequences are 7 and 3.
Katyanochek1 [597]

Answer:

1.1.1) -61

1.1.2) Term 42

Step-by-step explanation:

Tn= dn (difference) + t0 (term 0)

Tn= -4n+ 11

You can now take this general rule to find out the other terms.

Tn= -4(18)+11

Tn= -61

And the term -157

Tn= -4n+11

-157= -4n + 11

-157 - 11= - 4n

-168= -4n

-168/-4= n

42= n

3 0
3 years ago
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