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alex41 [277]
3 years ago
12

What is the following quotient? startfraction startroot 6 endroot + startroot 11 endroot over startroot 5 endroot + startroot 3

endroot endfraction.
SAT
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

its b

Explanation:

because i just took the test baddies

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A state highway department uses a salt storage enclosure that is in the shape of a cone, as shown above. If the volume of the st
denis-greek [22]

Answer:

D = 8 m

Explanation:

vol = πr² h/3

48π = πr² x 9/3

48π = 3πr²

r = sqrt (48/3)

r = 4 m

therefore the diameter D = 2r = 2 (4) = 8 m

3 0
4 years ago
What is the ph of a 0. 300 m solution of aniline (c₆h₅nh₂, kb = 4. 3 × 10⁻¹⁰)?.
yawa3891 [41]

The pH of the solution of aniline is; pH = 4.64

<h3>pH of aniline solution</h3>

According to the question;

  • We are required to determine the hydrogen ion concentration, pH of the given aniline solution.

Since, the given Kb is kb = 4. 3 × 10⁻¹⁰.

Hence, it follows that;

  • Ka = 10-¹⁴/4. 3 × 10⁻¹⁰.

  • Ka = 2.3 × 10⁻⁵.

On this note, the pH of aniline solution is;

  • pH = -log {2.3 × 10⁻⁵}

pH = 4.64.

Ultimately, the pH of the aniline solution given is; 4.64

Read more on pH of solutions;

brainly.com/question/25639733

3 0
2 years ago
What brought large amounts of water into the city of rome? a. canals b. forums c. bridges d. aqueducts please select the best an
skelet666 [1.2K]

Hello, hope this helps

Answer:

D. Aqueducts

Explanation:

They constructed these to bring water to many services such as public baths and latrines. Let me know if have any questions about this.

4 0
3 years ago
Factor the expression into an equivalent form 12y^2-75.
elena-14-01-66 [18.8K]

Hello oddworld7836!

\huge \boxed{\mathbb{QUESTION} \downarrow}

Factor the expression into an equivalent form 12y² - 75.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

12 y ^ { 2 } - 75

By observing the expression, we can see that, 3 is the only common factor in both the terms of the expression. So, take the common factor 3 out.

12 y ^ { 2 } - 75 \\  = 3\left(4y^{2}-25\right)

Now, look at (4y² - 25). They don't have any common factors but they appear in the form of the algebraic identity ⇨ a² - b² = (a + b) (a - b). Here,

  • a² = 4, a = 2 (√a² = ✓4 = 2)
  • b² = 25, b = 5 (√b² = ✓25 = 5)

So, the (4y² + 25) becomes...

(4 {y}^{2}  - 25) \\  = \left(2y-5\right)\left(2y+5\right)

Now, bring the 3 (common factor) & rewrite the complete expression.

12 y ^ { 2 } - 75 \\  =  \boxed{ \boxed{ \bf \: 3\left(2y-5\right)\left(2y+5\right) }}

We can't further simplify it. Also, remember that the simplified form of an expression is equivalent to the expression. So, 3 (2y - 5) (2y + 5) is equivalent to 12y² - 75.

__________________

Hope it'll help you!

ℓu¢αzz ッ

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