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3 years ago

(2a²b 5)2 (ab 332 22(a²b-c-3)(a?c%)

Mathematics

1 answer:

Gemiola [76]3 years ago

8 0

Gfdvhhygvfyhbfghvhfghgvvhhhbhhh

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If x< 0 and y<0 where is the point (x,y) located

Phoenix [80]

Answer: 0

Step-by-step explanation:

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3 years ago

3. What is the hypothesis of the following conditional statement?

lys-0071 [83]

Answer:

A: Two angles form a right angle.

Step-by-step explanation:

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3 years ago

Find the value of in the triangle shown below<br><br> 9, 7, and x around the triangle

enot [183]

Answer:

5.66 units

Step-by-step explanation:

A's per our discussion, x is a leg of right triangle.

Therefore, by Pythagoras theorem:

$x = \sqrt{ {9}^{2} - {7}^{2} } \\ = \sqrt{81 - 49} \\ = \sqrt{32}$

$x = \sqrt{ {16}^{2} - {12}^{2} } \\ = \sqrt{256 - 144} \\ = \sqrt{112}$

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3 years ago

Factor 9x^2+66xy+121y^2

Korvikt [17]

Answer: (3x + 11y)^2

Demonstration:

The polynomial is a perfect square trinomial, because:

1) √ [9x^2] = 3x

2) √121y^2] = 11y

3) 66xy = 2 *(3x)(11y)

Then it is factored as a square binomial, being the factored expression:

[ 3x + 11y]^2

Now you can verify working backwar, i.e expanding the parenthesis.

Remember that the expansion of a square binomial is:

- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2

=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.

7 0

4 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$

Comparing equations (A) and (B), we have

$5x-1=7x^2-1\\\\\Rightarrow 5x=7x^2\\\\\Rightarrow x=\dfrac{5}{7},~\textup{since }x\neq 0.$

Thus,

$AM=5\times\dfrac{5}{7}=\dfrac{25}{7}=3\dfrac{4}{7}.$

8 0

4 years ago