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3 years ago

Algebraically solve for the roots of the quadratic function written in vertex form found below

Mathematics

1 answer:

muminat3 years ago

6 0

Answer:maybe this work/ helps if not -sorry-

Step-by-step explanation:

equation :

y-(-2*(x+5)^2+18)=0

STEP

Equation at the end of step 1

y - ((0 - 2 • (x + 5)2) + 18) = 0

STEP

Equation at the end of step 2

y + 2x2 + 20x + 32 = 0

STEP

Solving a Single Variable Equation:

3.1 Solve y+2x2+20x+32 = 0

In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.

We shall not handle this type of equations at this time.

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Multiply the polynomials. (4x- + 4x + 6)(7x + 5)

SashulF [63]

$\implies {\blue {\boxed {\boxed {\purple {\sf { \: 28 {x}^{3} + 48 {x}^{2} + 62x + 30}}}}}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

$(4 {x}^{2} + 4x + 6)(7x + 5)$

$= \: 7x(4 {x}^{2} + 4x + 6) + 5(4 {x}^{2} + 4x + 6)$

$= \: 28 {x}^{3} + 28 {x}^{2} + 42x + 20 {x}^{2} + 20x + 30$

Combining like terms, we have

$= \: 28 {x}^{3} + (28 {x}^{2} + 20 {x}^{2} ) + (42x + 20x) + 30$

$= \: 28 {x}^{3} + 48 {x}^{2} + 62x + 30$

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}$

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3 years ago

The figure below shows square ABCD inscribed on square PQRM on a coordinate plane. Which of the following expressions represents

vova2212 [387]

The hypotenuse of the triangle is given by:
$CB ^ 2 = RC ^ 2 + RB ^ 2
$
Substituting values:
$CB ^ 2 = 2 ^ 2 + 4 ^ 2
$
Clearing CB we have:
$CB = \sqrt{ 2 ^ 2 + 4 ^ 2} 

 CB = \sqrt{ 4 + 16} 

 CB = \sqrt{20} 

 CB = \sqrt{4*5} 

 CB = 2\sqrt{5} 



$
Then, the perimeter of the triangle is:
$P = RB + RC + CB
$
Substituting values:
$P = 4 + 2 + 2 \sqrt{5} 

 P = 6 + 2 \sqrt{5}$
Answer:
the perimeter of triangle BRC in units is:
D.6 + 2√5

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Whoever answers this gets brainliest plsss help

Karolina [17]

hope it helped. I have solved in the same paper. plz don't mind.

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