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PolarNik [594]
2 years ago
5

Which is the completely factored form of 4x2 28x 49? (x 7)(4x 7) 4(x 7)(x 7) (2x 7)(2x 7) 2(x 7)(x 7)

Mathematics
1 answer:
natima [27]2 years ago
8 0

The completely factored form of 4x^2 + 28x + 49 is given by: Option C: (2x+7)(2x+7)

<h3>How to find the factors of a quadratic expression?</h3>

If the given quadratic expression is of the form ax^2 + bx + c, then its factored form is obtained by two numbers alpha( α ) and beta( β) such that:

b = \alpha + \beta \\ ac =\alpha \times \beta

Then writing b in terms of alpha and beta would help us getting common factors out.

Sometimes, it is not possible to find factors easily, so using the quadratic equation formula can help out without any  trial and error.

For this case, the given quadratic expression is:

4x^2 + 28x + 49

So we've to find two numbers such that:

Their sum = b =  28

Their product = ac = 4\times 49 = 196

We can see that 196 is square of 14, and that 14 added twice forms 28, thus:

14 + 14 = 28

14×14 = 196

Writing b = 28 as sum of 14 twice, we get:

4x^2 + 28x + 49 = 4x^2 + 14x + 14x +49 \\4x^2 + 28x + 49 =  2x(2x+7) + 7(2x + 7)\\4x^2 + 28x + 49  = (2x+7)(2x+7)

Thus, the completely factored form of 4x^2 + 28x + 49 is given by: Option C: (2x+7)(2x+7)

Learn more about factorization of quadratic expression here:

brainly.com/question/26675692

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tino4ka555 [31]

Given:

The inequality is:

3t+1\leq 5

To find:

The graph of the given inequality.

Solution:

We have,

3t+1\leq 5

Subtract both sides by 1.

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Divide both sides by 3.

t\leq \dfrac{4}{3}

The value of t is less than or equal to \dfrac{4}{3}.

Since t\leq \dfrac{4}{3}, it means \dfrac{4}{3} is included in the solution, therefore there is a  closed circle at t=\dfrac{4}{3} and an arrow approaches to left from t=\dfrac{4}{3}.

Therefore, the correct option is A.

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2 years ago
HELPP!!what is the answer to this picture above^
nasty-shy [4]

Answer:

\fbox{ \fbox { \sf{Distance  =  \sf{15 \: units}}}}

{ \fbox { \fbox { \sf{Midpoint = { \sf{ \: (12 \: , \: 10.5)}}}}}}

Step-by-step explanation:

\star{ \:  \sf { \: Let \: the \: points \: be \: A \: and \: B}}

\star { \sf{Let \: A(6 \:, 6) \: be \: (x1 ,\: y1) \: and \: B(18 ,\: 15) \: be \: (x2 \:, y2)}}

\underline{ \underline{ \tt{Finding \: the \: distance}}}

\boxed{ \sf{Distance =  \sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(18 - 6)}^{2}  +  {(15 - 6)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(12)}^{2}  +  {(9)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{144 + 81}}}

\mapsto{ \sf{Distance =  \sqrt{225} }}

\mapsto{ \sf{Distance =  \sqrt{ {15}^{2} } }}

\mapsto{  \boxed{\sf{Distance = 15 \: units}}}

\underline{ \underline {\tt{Finding \: the \: Midpoint}}}

\boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2}  \: , \:  \frac{y1 + y2}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{6 + 18}{2}  \: , \:  \frac{6 + 15}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{24}{2}  \: , \:  \frac{21}{2} )}}

\mapsto{ \boxed{ \sf{Midpoint = (12 \: , \: 10.5)}}}

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

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