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3 years ago

Solve the equation

Mathematics

1 answer:

Vikki [24]3 years ago

4 0

Answer:

x ≈ - 5.32, x ≈ 1.32

Step-by-step explanation:

Given

x² + 4x - 7 = 0 ( add 7 to both sides )

x² + 4x = 7

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(2)x + 4 = 7 + 4

(x + 2)² = 11 ( take the square root of both sides )

x + 2 = ± $\sqrt{11}$ ( subtract 2 from both sides )

x = - 2 ± $\sqrt{11}$

Thus

x = - 2 - $\sqrt{11}$ ≈ - 5.32 ( to 2 dec. places )

x = - 2 + $\sqrt{11}$ ≈ 1.32 ( to 2 dec. places )

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Choose one of the factors of 500x3 + 108y18

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Answer:

Option C

Therefore, one of the factors of $500x^3 +108y^\left (18\right )$ is $(5x+3y^6)$ .

Step-by-step explanation:

Given: $500x^3 +108y^\left (18\right )$

the common factor from $500x^3$ and $108y^\left (18\right )$ is 4.

therefore, $4\cdot \left ( 125x^3+27y^\left (18\right ) \right )$

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$( (5x)^3+(3y^6)^3 \right))$ $=(5x+3y^6)(25x^2-15xy^6+9y^12)$

Therefore, we have

$500x^3 +108y^\left (18\right )$ $=4\cdot (5x+3y^6)\left (25x^2-15xy^6+9y^\left ( 12 \right ) \right )$

Therefore, one of the factors of $500x^3 +108y^\left (18\right )$ is $(5x+3y^6)$ .

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3 years ago

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1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie

Ad libitum [116K]

Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
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U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

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Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21

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3 years ago

A jar contains $14.25 in quarters and dimes. There are 75 coins in the jar. Find the number of quarters in the jar.

Marianna [84]

Set up a system of equations.
d + q = 75
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Multiply the first equation by -.1 so that the "d's" will cancel.
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