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Sholpan [36]
3 years ago
15

(1-5q) - (2.5s + 8 ) - (0.5q + 6 )

Mathematics
2 answers:
Sladkaya [172]3 years ago
4 0
First rewrite it by distributing the negatives.
1-5q-2.5s-8-.5q-6
Now combine like terms.
-13-5.5q-2.5s
Aleksandr-060686 [28]3 years ago
3 0

Answer:

-5.5q-2.5s-13

Step-by-step explanation:

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The value of k is: <br><br> please help im struggling
iVinArrow [24]

Answer:

  • E) 48

Step-by-step explanation:

<em>Parallel lines divide the transversals proportionally.</em>

Set proportions and solve for k:

  • k/24 = 64/32
  • k/24 = 2
  • k = 24*2
  • k = 48

Correct choice is E.

4 0
2 years ago
Which of the following is an improper integral:
Alla [95]

Answer:

∫₀² ln(x²) dx

Step-by-step explanation:

An integral is an improper integral if one or both endpoints is infinity, if the function is undefined at one or both endpoints, or if the function is discontinuous between the endpoints.

ln(x²) is undefined at x = 0.

6 0
3 years ago
The ratio of girls to boys in a classroom is 5 to 4.how many girls are in the class if there is a total of 27 students
statuscvo [17]
5 girls : 4 boys
5 + 4 = 9 students

= 27 students ( 5girls / 9 students )
*cancel students units*
= 27 ( 5 girls / 9)
= 15 girls
4 0
3 years ago
What is the product in simplest form? State any restrictions on the variable. Is it 1,2,3 or 4
mart [117]

Answer: I think its C

Step-by-step explanation:

6 0
2 years ago
What is the area (in square units) of the region under the curve of the function f(x)=x+3, on the interval from x=1 to x=3 ?
aleksandrvk [35]

Answer:

10 square units

Step-by-step explanation:

We want to find the area under the curve f(x)=x+3 from x=1 to x=3.

We use definite integrals to find this area.

\int\limits^3_1 {x+3} \, dx

We integrate to obtain:

\frac{x^2}{2}+3x|_1^3

We evaluate the limits to get:

\frac{3^2}{2}+3(3)-(\frac{1^2}{2}+3(1))

4.5+9-0.5-3=10

Therefore the area under the curve from x=1 to x=3 is 10 square unit.

8 0
3 years ago
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