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2 years ago

Solve for x in the equation x squared 14 x 17 = negative 96.

Mathematics

1 answer:

Ilia_Sergeevich [38]2 years ago

8 0

The value of x in the given quadratic equation is determined as -7 ± 8i.

<h3>Solution of the quadratic equation</h3>

The solutiion to the linear equation is determined as follows;

x² + 14x + 17 = -96

x² + 14x + 17 + 96 = 0

x² + 14x + 113 = 0

solve the equation using formula method;

a = 1, b = 14, c = 113

$x = \frac{-b \ \ \pm\sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{- 14\ \ \pm\sqrt{(14)^2 - 4(1\times 113)} }{2(1)} \\\\x = \frac{- 14\ \ \pm\sqrt{-256} }{2}\\\\x = \frac{- 14\ \ \pm\sqrt{256} \times \sqrt{-1} }{2}\\\\x = \frac{-14 \ \ \pm 16 \times i}{2} \\\\x = -7 \pm 8i$

Thus, the value of x in the given quadratic equation is determined as -7 ± 8i.

Learn more about quadratic equation here: brainly.com/question/1214333

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lisabon 2012 [21]

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3 years ago

Suppose f is an exponential function and:

Nastasia [14]

Answer: a) $b= p(1 + r)^1$ b) $c= p(1 + r)^n$ c) $d= p(1 + r)^m$ d) $b^n=c$

Step-by-step explanation:

Since f is an exponential function.

So, it is expressed as

$f = p(1 + r)^$

Here f is the present value

p is the initial value

r is the rate of growth per time period

t is the time period.

(a) b represents the 1-unit growth factor for f.

$b= p(1 + r)^1$

(b) c represents the n n-unit growth factor for f.

$c= p(1 + r)^n$

$d= p(1 + r)^m$

Write a formula that expresses c in terms of b.

Since

$b=p(1+r)^1\\\\So,\\\\b=p(1+r)^{n\times \frac{1}{n}}\\\\b=(p(1+r)^n)^{\frac{1}{n}}\\\\b=c^{\frac{1}{n}}\\\\b^n=c$

Hence, a) $b= p(1 + r)^1$ b) $c= p(1 + r)^n$ c) $d= p(1 + r)^m$ d) $b^n=c$

8 0

3 years ago

Please help with 2 questions. Thank you.

8_murik_8 [283]

${e}^{ln(anything)} = anything$

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${e}^{ln1} = 1$

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8 0

3 years ago

I need to send photos.

frez [133]

SOLUTION:

Step 1:

From the given question, and comparing the scores in classes A and B

Step 2:

Question one, to know the class which had the better overall result on the exam;

(I) The scores in class A ranges from 65 to 100, while that of class B ranges from 60 to 90. This explains that class A had better results.

(ii) The median of the scores in class A is 85, while the median of the scores in class B is 75, this is another piece of supporting evidence that class A had better results.

Step 3:

Question two, to know the class which had greater variability in the results;

For a better understanding of this question, I need to explain the concept of variability in statistics.

Variability in statistics refers to the difference being exhibited by data points within a data set, as related to each other or as related to the mean. This can be expressed through the range, variance or standard deviation of a data set.

Step 4:

So we need to find the range of scores in each of the two classes and then compare, the class with the greater range has the greater variability.

The Range is the difference between the lowest and highest score (H - L), where H is the highest score and L is the lowest score.

Step 5:

Applying the formula for range stated in step 4;

For Class A; H = 100 and L = 65

For Class B; H = 90 and L = 60

The range for class A; H - L = 100 - 65 = 35

The range for class B; H - L = 90 - 60 = 30

By comparing the range of class A and that of B, it is clear that Class A had a greater range (variability)

CONCLUSION:

Class A had better overall results in the exam and greater variability in the results.

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1 year ago

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balandron [24]

The answer is 8.7 cm

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3 years ago

Read 2 more answers