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lutik1710 [3]
3 years ago
8

Heidi

Mathematics
1 answer:
ivolga24 [154]3 years ago
7 0

Answer:

yes

Step-by-step explanation:

ok

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What is the slope of a line parallel to the line with equation 5x 3y=7?
adell [148]
-5/3
down 5, over 3
:)
6 0
3 years ago
Correct answer get the BRAINLIEST ANSWER!
OLga [1]
Here is how I completed the problem:

I started by creating a ratio:

\frac{x}{0.18} + \frac{360}{0.10} = \frac{x+360}{0.15}

<em>Where 'x' is the desired amount of the 18% solution of sulfuric acid. 

</em>From here, I solved for 'x':
<em>
</em>\frac{0.10x}{0.018} + \frac{360(0.18)}{0.018} = \frac{x+360}{0.15}
\frac{0.10x}{0.018} + \frac{64.8}{0.018} = \frac{x+360}{0.15}
\frac{0.10x+64.8}{0.018} = \frac{x+360}{0.15}
0.15(0.10x+64.8) = 0.018(x+360)
0.015x+9.72 = 0.018x+6.48
0.003x = 3.24
x = 1080

∴You will need to add 1080ml of 18% Sulfuric Acid in order to obtain a 15% solution.
8 0
3 years ago
Please solve with explanation
OverLord2011 [107]

Real life scenarios of acute angles are:

  • Sighting a ball from the top of a building at an angle of 55 degrees.
  • The angle between two adjacent vanes of a fan that has 6 vanes

<h3>What are acute angles?</h3>

As a general rule, an acute angle, x is represented as: x < 90

This means that acute angles are less than 90 degrees.

<h3>The real life scenarios</h3>

The real life scenarios that involve acute angles are scenarios that whose measure of angle is less than 90 degrees.

Sample of the real life scenarios that satisfy the above definition are:

  • Sighting a ball from the top of a building at an angle of 55 degrees.
  • The angle between two adjacent vanes of a fan that has 6 vanes

Read more about acute angles at:

brainly.com/question/3217512

#SPJ1

8 0
2 years ago
The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr
Kobotan [32]

Answer:

%  change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

T = 2.5 x + 0.5 x^{2}

We will approximate this distance  using the relation

f (x + dx) = f (x)+ f' (x)dx

dx = 26 - 25 = 1

T' =  2.5 + x

Therefore

f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× 25^{2} + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × 25^{2}

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

%  change in stopping distance = \frac{27.5}{375} × 100

%  change in stopping distance = 7.34 %

6 0
3 years ago
What is system of equations for (6,5)
worty [1.4K]

Answer:

no clue

Step-by-step explanation:

4 0
3 years ago
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