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ankoles [38]
3 years ago
8

Is proportional or not?

Mathematics
1 answer:
Eduardwww [97]3 years ago
8 0

top four are yes yes no yes

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A drilling crew dug to a depth of 26 ½ feet during their first day of drilling. On the second day, the crew dug down 9½ feet mor
Neporo4naja [7]

The depth of the bottom of the hole after the second day is 36 feet using addition operation.

<h3>What is addition?</h3>

In math, addition is the process of adding two or more integers together. Addends are the numbers that are added, while the sum refers to the outcome of the operation.

Given the depth on the first day is 26 ½ feet.

Depth on the second day = 9½ feet more than on the first day i.e. 9½ feet + depth on the first day

This implies, depth on the second day = 9½ + 26 ½

= 36 feet

Therefore, the depth of the bottom of the hole after the second day is 36 feet.

To learn more about addition, visit:

brainly.com/question/25621604

#SPJ9

5 0
1 year ago
11 3<br>- 50 ml =<br>A 2.41<br>B 2.91<br>C<br>3.351<br>D<br>3.451​
julsineya [31]
Es número c por qué esa es la respuesta
5 0
4 years ago
A triangle has side lengths of 4, inches, 10 inches, and c inches.
alexdok [17]

Answer:

c<4+10

c<14

Step-by-step explanation:

can I get brainliest?

7 0
3 years ago
Read 2 more answers
What is the value of x? (5 points)
Svetlanka [38]
30 + 2x = 90
2x = 60
x = 30

Answer:
x = 30
7 0
3 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
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