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vaieri [72.5K]
3 years ago
15

What is the value of the expression 3x+4 when x=7?

Mathematics
2 answers:
Pani-rosa [81]3 years ago
7 0

Answer: 25

Step-by-step explanation: 7x3 = 21+4 =25

Hope this helped!

Mark Brainliest if you want!

lara31 [8.8K]3 years ago
7 0

Answer:

25

Step-by-step explanation:

3x +4

3×7+4

21 +4

25

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spayn [35]

(6y-2y+5y)9

= 54y-18y+45y

80-10y

=10(8-y)

5 0
2 years ago
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Subtract 16 L 18 CL 9 ml from 85 L
harkovskaia [24]

Answer:

  (1)  68.811 L

Step-by-step explanation:

The amount being subtracted is ...

  16 + 0.18 + 0.009 = 16.189 . . . . liters

The difference is ...

  85.000 -16.189 = 68.111 . . . liters

__

<em>Additional comment</em>

cL is the SI unit "centiliters." The prefix "centi-" mean 1/100 or 0.01 units. Then 18 cL = 18 × 0.01 L = 0.18 L.

mL is the SI unit "milliliters." The prefix "milli-" means 1/1000 or 0.001 units. Then 9 mL = 9 × 0.001 L = 0.009 L.

8 0
2 years ago
Find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.) dx x(l
Pepsi [2]

I'm assuming the integral is

\displaystyle \int \frac{dx}{x (\ln(x^2))^5}

We have

\ln(x^2) = 2 \ln|x| \implies (\ln(x^2))^5 = 32 (\ln|x|)^5

Then substituting y=\ln|x| and dy=\frac{dx}x, the integral transforms and reduces to

\displaystyle \int \frac{dx}{x(\ln(x^2))^5} = \frac1{32} \int \frac{dy}{y^5} \\\\ ~~~~~~~~ = \frac1{32} \left(-\frac1{4y^4}\right) + C \\\\ ~~~~~~~~ = -\frac1{128(\ln|x|)^4} + C

which we can rewrite as

128 (\ln|x|)^4 = 8\cdot2^4(\ln|x|)^4 = 8 (2\ln|x|)^4 = 8 (\ln(x^2))^4

and so

\displaystyle \int \frac{dx}{x (\ln(x^2))^5} = \boxed{-\frac1{8(\ln(x^2))^4} + C}

7 0
1 year ago
Help please! This really impacts my grade!! Hope you guys know the answers
Leokris [45]

Answer: this dose not even make sense

Step-by-step explanation:

You only put a pdf

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3 years ago
Identify the domain and range
UkoKoshka [18]

Answer:

Domain: 1 ≤ x ≤ 4

Range : 1 ≤ f(x) ≤ 4

Step-by-step explanation:

The domain of a function f(x) is the limit within which the values of x varies.

Here, in the graph, it shows that the maximum value of x is 4 and the minimum value of x is 1.

Therefore, the domain of the function is 1 ≤ x ≤ 4

Again the range of a function f(x) is the limit within which the values of f(x) vary.

Here, the graph shows that the maximum value of f(x) is 4 and the minimum value of f(x) is 1.

Therefore, the range of the function f(x) is 1 ≤ f(x) ≤ 4. (Answer)

6 0
3 years ago
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