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3 years ago

If x2 - 4 = 45, then x could be equal to 9 3 5 -7 -4

1 answer:

5 0

Transpose the constant on the left hand side of the equation to the right side. The equation becomes x² = 45 + 4,
x² = 49
x = sqrt(49)
x = 7 and x = -7
Only -7 is on the choices. Thus, the answer is x = -7.

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Use slopes and y-intercepts to determine if the lines 5x-5y=-2 and -x+2y=4

HACTEHA [7]

The given pair of lines are not perpendicular.

The line is a curve showing the shortest distance between 2 points.

5x - 5y = -2 - - - - - (1)
Transform the equation into standard form,
5x + 2 = 5y
y = 5x /5 + 2/5
y = x + 2/5

The slope of equation 1 is $m_1 = 1$ and intercept c = 2 / 5

Similarly
x + 2y = 4 - - - - - - - -(2)
Transform it into standard form
y = -x/2 + 4 /2
y = -x / 2 + 2

Slope of the equation 2 $m_2$ = -1 / 2 and intercept c = 2
Slope of line 1 * slope of line 2 = 1 * -1/2 = -1/2

Since the lines are not perpendicular because the pair of lines does not satisfy the property of perpendicular lines i.e
$m_1*m_2 = -1$

Thus, the given pair of lines are not perpendicular.

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6 0

1 year ago

What is 3/4-(-2/9) I need hekppp

Sonbull [250]

Answer:

35/36

Step-by-step explanation:

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3 0

3 years ago

Erica has eight necklaces 1/4 of the necklaces are blue how many necklaces are blue ?

timofeeve [1]

Two of the necklaces are blue

7 0

3 years ago

Read 2 more answers

Factor completely: <br> x^3 - x^2 - 6x

PSYCHO15rus [73]

Answer:

$x(x-3)(x+2)$

Step-by-step explanation:

1. Find the GCF (Greatest Common Factor)

$GCF=x$

2. Factor out the GCF and simplify.

$x(x^{2} -x-6)$

3. Factor $x^{2} -x-6.$

Which two numbers add up to -1 and multiply to -6?

Rewrite the expression using the numbers above.

$(x-3)(x+2)$

4. Done!

$x(x-3)(x+2)$

4 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago