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3 years ago

Solve using the quadratic formula. Show all work. Write each solution in simplest form. No decimals.

Mathematics

1 answer:

alexira [117]3 years ago

3 0

Answer:

Option B

Step-by-step explanation:

Given quadratic equation is,

12a² + 9a + 7 = 0

By comparing this equation with standard quadratic equation,

hx² + kx + c = 0

h = 12, k = 9 and c = 7

By using quadratic formula,

a = $\frac{-k\pm\sqrt{k^2-4hc}}{2h}$

= $\frac{-9\pm\sqrt{9^2-4(12)(7)}}{2(12)}$

= $\frac{-9\pm\sqrt{81-336}}{2(12)}$

= $\frac{-9\pm\sqrt{-255}}{24}$

= $\frac{-9\pm i\sqrt{255}}{24}$

a = $\frac{-9+ i\sqrt{255}}{24},\frac{-9- i\sqrt{255}}{24}$

Therefore, Option B will be the correct option.

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The real root of the equation x3 – 7x2 + 15x – 25 = 0 is 5. What are the nonreal roots?

Elodia [21]

Answer:

1±2i.

Step-by-step explanation:

1. if to evaluate the expression given, it can be written

(x-5)(x²-2x+5)=0;

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2 years ago

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irina1246 [14]

Answer:

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Step-by-step explanation:

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3 years ago

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What are 3 numbers to multiply to equal 750?

Rainbow [258]

75 * 5 * 2 = 750

150 * 5 * 1 = 750

375 * 2 * 1 = 750

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3 years ago

Julissa is running a 10-kilometer race at a constant pace. After running for 18 minutes, she completes 2 kilometers. After runni

IRINA_888 [86]

So18 is to 2 and54 is to 6
18:2 and 54:6
18/2 and 54/6
simplify
9/1 and 9/1
she is running at the same rate

so we need to do
t times something=k
its eems that she divided the time by 9
so
1/9t=k

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4 years ago

An automobile dealer had 288 cars and trucks in stock during the month. He must pay an inventory fee of 52 per car and $5 per tr

Aleonysh [2.5K]

For the given conditions, we won’t get a practical solution.

<u>Solution:</u>

Given, An automobile dealer had 288 cars and trucks in stock during the month.

He must pay an inventory fee of 52 per car and $5 per truck.

He paid $849 for inventory foes.

We have to find how many cars and Trucks did he have during the month?

Now, let the number of trucks be n, then number of cars will be 288 – n

And, according to given information,

$\begin{array}{l}{5 \times n+52 \times(288-n)=849} \\\\ {\rightarrow 5 n+52 \times 288-52 n=849} \\\\ {\rightarrow 52 n-5 n=14976-849} \\\\ {\rightarrow 47 n=14127} \\\\ {\rightarrow n=300.57}\end{array}$

Here we got n > number of cars and trucks, which is practically impossible.

Hence, for the given conditions, we won’t get a practical solution

8 0

3 years ago