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3 years ago

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Determine the correct set up for solving the equation using the quadratic formula.

1 answer:

netineya [11]3 years ago

5 0

Answer: B

Begin by simplifying the given equation, so it's rearranged into standard form. Move all the terms onto one side, so they equal 0. This would result with $4x^{2}$ -3x-9=0. Quadratic formula is (-b±√(b²-4ac))/(2a). Standard form follow this variable format: $ax^{2}$ +bx+c. Use the numbers that correspond with each variable to substitute them into the quadratic formula. This would be

(-(-3)±√(-3²-4(4)(-9)/2(4).

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The area of a trapezoid is 60 square feet its height is 6 feet and one base is 12 feet.Find the number of feet in the other base

adell [148]

<h2>Greetings</h2>

Answer:

The other base is 8 feet.

Step-by-step explanation:

You need to recall the area of a trapezoid equation:

$\frac{a + b}{2} * h$

We need to work back, so we need to divide the area by 6:

60 ÷ 6 = $\frac{a + b}{2}$ = 10

Then we need to multiply this by two to find a + b value:

10 x 2 = a + b

20 = a + b

Beause we know the value of a, which is 12, we can factor this in:

20 = 12 + b

Move the +12 over to the other side, making it a negative:

20 - 12 = b

8 = b

<h3>So the other base is 8 feet!</h3>
<h2>Hope this helps!</h2>

5 0

3 years ago

This is due like right now, please help me!!!!

Varvara68 [4.7K]

$\bold{\huge{\orange{\underline{ Solution }}}}$

$\bold{\underline{ Given \: Rules}}$

<u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
<u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
<u>The </u><u>sum </u><u>of </u><u>any </u><u>row </u><u>does </u><u>not </u><u>equal </u><u>the </u><u>sum </u><u>of </u><u>any </u><u>column </u>

$\bold{\underline{ Let's \: Begin}}$

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

$\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }$

$\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}$

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

$\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}$

$\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}$

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

$\sf{ 158 + b = 166 + a, 156 + d = 166 + a }$

$\sf{ b = 166 - 158 + a, d = 166 - 156 + a }$

$\sf{ b = 8 + a, d = 10 + a ...(3)}$

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{229+a = 161+8+a+c = 168+10+a+e}$

$\sf{ 229+a = 169+a+c = 178+a+e}$

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

$\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}$

$\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}$

$\sf{ c = 60 \: and \: e = 51 }$

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>c </u><u>and </u><u>e </u><u>in </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

$\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }$

$\sf{ 158 + b = 156 + d = 166 + a = 189}$

<u>Now</u><u>, </u>

$\sf{ For \: b, 158 + b = 189 }$

$\sf{ b = 189 - 158 }$

$\sf{ b = 31}$

$\sf{ For \: d , 156 + b = 189 }$

$\sf{ d = 189 - 156 }$

$\sf{ d = 33}$

$\sf{ For \: a, 166 + a = 189 }$

$\sf{ a = 189 - 166 }$

$\sf{ a = 23 }$

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

8 0

3 years ago

Whitepunk [10]

Here is the solved answer for this question

7 0

3 years ago

Fed [463]

Answer:

b -17

Step-by-step explanation:

−a^2 − 3b^3 + c^2 + 2b^3 − c^2 =

= -(3²) - 3(2³) + (-3)² + 2(2³) - (-3)²

= -9 - 3(8) + 9 + 2(8) - 9

= -9 - 24 + 9 + 16 - 9

= -17

8 0

2 years ago

What kind of sequence is this?<br> 5, 8, 13, 21, <br> arithmetic<br> geometric<br> both<br> neither

Ivanshal [37]

Answer:

Hi how are you doing today Jasmine

8 0

3 years ago