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3 years ago

Use the elimination method to solve the system of equations. Choose the correct ordered pair.

Mathematics

2 answers:

Vlad1618 [11]3 years ago

8 0

Answer: A

Step-by-step explanation:

kherson [118]3 years ago

6 0

Make x=0 in the second equation, giving you y=-3. If you plug that in to the top equation you would get x-3(-3)=-19, x+9=-19,x=-28.

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6.75 + StartFraction 3 Over 8 EndFraction x = 13 and one-fourth

amid [387]

Easy the answer is B: 17 and one-third

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3 years ago

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How to write an equation in logarithmic form?

laiz [17]

Answer:

here is a example

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3 years ago

What are the x- and y- coordinates of point E, which partitions the directed line segment from A to B into a ratio of 1:2? x = (

Contact [7]

Answer:

(0, 1)

Step-by-step explanation:

In the picture attached, the question is shown.

From the picture:

x2 = -4 (x-coordinate of the endpoint)

y2 = 9 (y-coordinate of the endpoint)

x1 = 2 (x-coordinate of the startpoint)

y1 = -3 (y-coordinate of the startpoint)

Ratio m:n, means that m = 1 and n = 2.

Replacing into the formulas:

x-coordinate = m/(m+n) * (x2- x1) + x1 = 1/3 * (-4 - 2) + 2 = 0

y-coordinate = m/(m+n) * (y2- y1) + y1 = 1/3 * (9 - (-3)) + (-3) = 1

6 0

2 years ago

A random sample of 145 students is chosen from a population of 4,250 students. The mean IQ in the sample is 130, with a standard

snow_tiger [21]

Answer:

Approximately, the 90% confidence interval for the students' mean IQ score is between 129.045 - 130.956

Step-by-step explanation:

The formula to use to solve this question is called the Confidence Interval formula.

Confidence interval =

x ± z × ( σ/ (√n) )

Where:

x = the sample mean = 130

z = the z-value for 90% confidence = 1.645

σ = standard deviation = 7

n = sample size = 145

130 ± 1.645 × (7/√145)

130 ± 0.9562687005

130 - 0.9562687005 = 129.0437313

130 + 0.9562687005 = 130.9562687005

Therefore, approximately, the 90% confidence interval for the students' mean IQ score is between 129.045 - 130.956

7 0

2 years ago

Evaluate cosA/2 given cosA=-1/3 and tanA >0

mote1985 [20]

Answer:

$\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}$

Step-by-step explanation:

Given that:

$cosA=-\dfrac{1}3$

and

$tanA > 0$

To find:

$cos\dfrac{A}{2} = ?$

Solution:

First of all,we have cos value as negative and tan value as positive.

It is possible in the 3rd quadrant only.

$\dfrac{A}{2}$ will lie in the 2nd quadrant so $cos\dfrac{A}{2}$ will be negative again.

Because Cosine is positive in 1st and 4th quadrant.

Formula:

$cos2\theta =2cos^2(\theta) - 1$

Here $\theta = \frac{A}{2}$

$cosA =2cos^2(\dfrac{A}{2}) - 1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =cosA+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =-\dfrac{1}3+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =\dfrac{2}3\\\Rightarrow cos(\dfrac{A}{2}) = \pm \dfrac{1}{\sqrt3}$

But as we have discussed, $cos\dfrac{A}{2}$ will be negative.

So, answer is:

$\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}$

7 0

2 years ago