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Arlecino [84]
3 years ago
5

Y-2=-3(x+4)2 in standard form

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

The standard for would be y = -3x^2 - 24x - 46

Step-by-step explanation:

To get this to standard form, start by squaring the parenthesis.

y - 2 = -3(x + 4)^2

y - 2 = -3(x^2 + 8x + 16)

Now distribute the -3

y - 2 = -3x^2 - 24x - 48

Now add the 2 to both sides

y = -3x^2 - 24x - 46

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72 miles in one day is 72 miles per how many hours
IrinaVladis [17]

Answer:

72 miles per 24 hours

Step-by-step explanation:

72 miles in one day

There are 24 hours in a day.

8 0
3 years ago
Which trigonometric ratio represents Cosine (cos)?
kotykmax [81]
If you want to remember the trig formulas, I suggest the “SOHCAHTOA” or the “Some old hippie caught another hippie tripping on acid method.” It combines all three principle formulas together. So, SOH means sine equals opposite over hypotenuse, and CAH means cosine equals adjacent over hypotenuse, and TOA means tangent equals opposite over adjacent.

Cosine equals adjacent/hypotenuse.
4 0
4 years ago
How do you solve 2. 310x/4y times 8y/5x =
Katyanochek1 [597]

Answer: 0.924

Step-by-step explanation:

Given

\frac{2.310x}{4y} *\frac{8y}{5x}

Simplify  \frac{2.310x}{4y} to \frac{0.5775x}{y}


\frac{0.5775x}{y}*\frac{8y}{5x}

Cancel y

0.5775x* \frac{8}{5x}

Cancel x

0.5775* \frac{8}{5}

Use the rule, a*\frac{b}{c} =\frac{ab}{c}

\frac{0.5775*8}{5}

Simplify

\frac{4.62}{5}

Simplify

0.924

6 0
2 years ago
The bisectors BO and CO of the angles at B and C of ABC meet at O. Prove that BOC = 90° + A/2.​
Maksim231197 [3]

The prove that BOC = 90° + A/2. is given below:

<h3>What is the bisector about?</h3>

Note that, bisectors of angles B and C and that of a triangle ABC is one that pass across each other at the point O.

Hence: one need to to prove that ∠BOC = 90° + A/2

Since:

BO is said to be the bisector of angle B e.g. ∠OBC = (∠1)

CO is said to be the bisector of angle C e.g.  ∠OCB = (∠2)

Looking at the triangle BOC, to interpret  by the use of angle sum property, then:

∠OBC + ∠BOC + ∠OCB = 180°

∠1 + ∠BOC + ∠2 = 180° ---- ( equation 1)

Looking at triangle ABC and by the use of angle sum property, So;

∠A + ∠B + ∠C = 180°

∠A + 2(∠1) + 2(∠2) = 180°

Then we divide by 2 on the two sides, and it will be:

∠A/2 + ∠1 + ∠2 = 180°/2

∠A/2 + ∠1 + ∠2 = 90°

∠1 + ∠2 = 90° - ∠A/2 ---(equation 2)

Then one need to Substitute (2) in (1), So:

∠BOC + 90° - ∠A/2 = 180°

∠BOC - ∠A/2 = 180° - 90°

∠BOC - ∠A/2 = 90°

Hence , ∠BOC = 90° + ∠A/2

Therefore, from the above, we care able to prove that BOC = 90° + A/2.

Learn more about bisector from

brainly.com/question/11006922

#SPJ1

3 0
2 years ago
Show that if x, y, z are integers such that x^3+5y^3 = 25z^3, then x = y = 0.
Gre4nikov [31]

Answer:

3125*k^9 + y^3 is an integer my closure property.

 but 5^(1/3) is not an integer, which forces z to be irrational.

Note that there is no way an integer value can rationalize 5^(1/3)

Step-by-step explanation:

x^3 = 25z^3 - 5y^3

x^3 = 5 ( 5z^3 - y^3)

x  =  (5 ( 5z^3 - y^3) )^(1/3) must be an integer

  = 5^(1/3) * (5z^3 - y^3)^(1/3)

Then  (5z^3 - y^3)^(1/3) = 25*k^3 for some integer k

 5z^3 - y^3 = 15625*k^9

 5z^3 = 15625*k^9 + y^3

  z^3 = 3125*k^9 + (1/5)*y^3

z =  ( 3125*k^9 + (1/5)*y^3 )^ ( 1/3)

4 0
3 years ago
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