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gizmo_the_mogwai [7]

3 years ago

5

Find the perimeter of a rectangle whose lenth is 24cm and diagonal is 25cm

1 answer:

VikaD [51]3 years ago

8 0

Answer:

using Pythagoras theorem

breadth = √25^2 -24^2

=√47

= 7cm

P = 24+7+24+7 = 62

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The angles of a triangle are 2x, 3x, and 4x degrees. Find the value of x.

dimulka [17.4K]

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3 years ago

Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

BaLLatris [955]

Answer:

$Vertex = (\frac{1}{2},\frac{7}{4})$

Step-by-step explanation:

Given

$f(x) = x^2 - x +2$

Required

The vertex

We have:

$f(x) = x^2 - x +2$

First, we express the equation as:

$f(x) = a(x - h)^2 +k$

Where

$Vertex = (h,k)$

So, we have:

$f(x) = x^2 - x +2$

--------------------------------------------

Take the coefficient of x: -1

Divide by 2: (-1/2)

Square: (-1/2)^2

Add and subtract this to the equation

--------------------------------------------

$f(x) = x^2 - x +2$

$f(x) = x^2 - x + (-\frac{1}{2})^2+2 -(-\frac{1}{2})^2$

$f(x) = x^2 - x + \frac{1}{4}+2 -\frac{1}{4}$

Expand

$f(x) = x^2 - \frac{1}{2}x- \frac{1}{2}x + \frac{1}{4}+2 -\frac{1}{4}$

Factorize

$f(x) = x(x - \frac{1}{2})- \frac{1}{2}(x - \frac{1}{2})+2 -\frac{1}{4}$

Factor out x - 1/2

$f(x) = (x - \frac{1}{2})(x - \frac{1}{2})+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+2 -\frac{1}{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{8 -1 }{4}$

$f(x) = (x - \frac{1}{2})^2+ \frac{7}{4}$

Compare to: $f(x) = a(x - h)^2 +k$

$h = \frac{1}{2}$

$k = \frac{7}{4}$

Hence:

$Vertex = (\frac{1}{2},\frac{7}{4})$

8 0

3 years ago

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Which polynomial correctly combines the like terms and puts the given polynomial in standard form? -5x^3y^3+8x^4y^2-xy^5-2x^2y^4

victus00 [196]

Given:

The polynomial is:

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

To find:

The simplified form of the given polynomial in standard form.

Solution:

We have,

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

Combining the like terms, we get

$=-5x^3y^3+8x^4y^2+(-2x^2y^4+3x^2y^4)+8x^6+(-xy^5-6xy^5)$

$=-5x^3y^3+8x^4y^2+x^2y^4+8x^6-7xy^5$

Now, rewrite this polynomial in standard form.

$=8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5$

Therefore, the required polynomial in standard form is $8x^6+8x^4y^2-5x^3y^3+x^2y^4-7xy^5$ .

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I am pretty sure it is A. Novels

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zvonat [6]

I'm gonna say that's a false accusation

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3 years ago