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Svet_ta [14]
3 years ago
11

Plzzz help

Mathematics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:C

Step-by-step explanation:

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rodikova [14]
U do 5 times 30 and get 150 then take 15 times 50 and get 750 last take 750 plus 150 and u get 900 cm
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What is the independent and dependent variable in this situation?
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Cid Sam is a way to remember
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8 0
4 years ago
The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?
egoroff_w [7]

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola.  The first few steps are as follows.  Set the parabola equal to 0 so you can solve for the vertex.  Separate the x terms from the constant by moving the constant to the other side of the equals sign.  The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out).  Let's start there.  The first 2 steps result in this polynomial:

-2x^2+4x=-1.  Now we factor out the -2:

-2(x^2-2x)=-1.  Now we complete the square.  This process is to take half the linear term, square it, and add it to both sides.  Our linear term is 2x.  Half of 2 is 1, and 1 squared is 1.  We add 1 into the set of parenthesis.  But we actually added into the parenthesis is +1(-2).  The -2 out front is a multiplier and we cannot ignore it.  Adding in to both sides looks like this:

-2(x^2-2x+1)=-1-2.  Simplifying gives us this:

-2(x^2-2x+1)=-3

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex.  Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

-2(x-1)^2+3=y

From this form,

y=-a(x-h)^2+k

you can determine the coordinates of the vertex to be (1, 3)

5 0
3 years ago
Read 2 more answers
If p is a positive integer, which could be an odd integer?
sdas [7]
p > 0 \Rightarrow

2p+2 (even)

p^3-p (odd or even)

p^2 + p (even)

p^2 - p (even)

7p - 3 (ODD); 7p odd, 7p-3 = (odd)-(odd)=(odd)
3 0
3 years ago
Read 2 more answers
13.......................
Alla [95]

Answer:

The answer is (d) ⇒ pq^{2}r\sqrt[3]{pr^{2}}

Step-by-step explanation:

* To simplify the cube roots:

 If its number then the number must be written in the form x³

 then we divide the power by 3 to cancel the radical

 If its variable we divide its power by 3 to cancel the radical

∵ \sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}

∴ p^{\frac{4}{3}}q^{2}}r^{\frac{5}{3}}=p^{1\frac{1}{3}}q^{2}r^{1\frac{2}{3}}

∵ p^{\frac{1}{3}}=\sqrt[3]{p}

∵ r^{\frac{2}{3}}=\sqrt[3]{r^{2}}

∴ p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})

∴ prq^{2}\sqrt[3]{pr^{2}}}

∴ The answer is (d)

5 0
3 years ago
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