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Soloha48 [4]
3 years ago
14

Can you help me on 34 and 35 please

Mathematics
1 answer:
dusya [7]3 years ago
6 0

-- #34 is choice-C .

-- See the picture attached to this answer, for both solutions.

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DO NOT SEND ME PDFS OR I WILL REPORT ANSWER CORRECTLY FOR BRAINLIEST
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Answer:

D is the answer

Step-by-step explanation:

Because everyone wants smart phones so just make them all smart phones lol

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3 years ago
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The polynomial below is perfect square trinomial of the form a^2 -2AB +B^2
guajiro [1.7K]

Answer:

In order to get this perfect square trinomial, the factored form will be (a - b)^2

Step-by-step explanation:

We know this because it follows the rule of perfect squares in which the ends are both squares and the middle term is -2 times each term.

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3 years ago
In triangle MNP, m = 5 cm, n = 4 cm, and p = 8 cm. Which formula can you use to find P?
charle [14.2K]
Answer:
8² = 4² + 5² - 2(4)(5) cos(P)
which is the third option

Explanation:
The general rule of cosine is:
a² = b² + c² - 2bc*cos(A)

In our triangle:
a is substituted by p = 8 cm
b is substituted by n = 4 cm
c is substituted by m = 5 cm
A is substituted by P which we want to find

Replace the variables in the general equation with the givens as follows:
p² = n² + m² - 2mn*cos(P)
8² = 4² + 5² - 2(4)(5) cos(P)

Hope this helps :)
4 0
3 years ago
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H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions?
zubka84 [21]
H(x)=(x^2)+1\ \ \ \ and\ \ \ \  K(x)=-(x^2)+4. \\\\K(H)=-(H^2)+4=-(x^2+1)^2+4=4-(x^2+1)^2=\\\\.\ \ \ =2^2-(x^2+1)^2=(2-x^2-1)(2+x^2+1)=(1-x^2)(3+x^2)=\\\\.\ \ \ =(1-x)(1+x)(x^2-3\cdot i^2)=(1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)\\\\ K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)=0\\\\x=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=\sqrt{3} \cdot i\ \ \ \ or\ \ \ \ x=-\sqrt{3} \cdot i\\\\Ans.\ e.
4 0
3 years ago
Can someone help me please
madreJ [45]

Answer:

2<x<8

Here is the answer to he above question you asked. (I have to type at least 20 words to post my answer...)

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3 years ago
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