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alex41 [277]
3 years ago
11

What is the solution to the system of equations? {x=5y=2x−1 (5, 9) (9, 5) (5, 11) (11, 5)

Mathematics
2 answers:
Maksim231197 [3]3 years ago
8 0

Answer:

ooga

Step-by-step explanation:

booga

stepladder [879]3 years ago
6 0

Answer:

(5, 9)

Step-by-step explanation:

Given:

The equations to solve are:

x=5\\y=2x-1

Since the value of 'x' is given, we solve this system of equations using Substitution method.

We have to solve this by substituting in the value of 'x' from first equation into second equation.

Therefore, the second equation becomes

y=2(5)-1\\\\y=10-1\\\\y=9

Therefore, the values are: x=5, y=9

Hence, the solution is in ordered pair form is (5, 9).

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The answer to this is B
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Oliver needs to save at least $1500 to buy a computer. He has already saved $650. How much more does he need to save? Write and
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X + 650 = 1500
       x     =  1500 - 650
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What is the value of x in the equation 2^2x = 2^3?
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Answer:

B


Step-by-step explanation:


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Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1
sattari [20]
We have that

N 1)
a²<span> − 9a + 14 = 0 
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

a2=4.5-√6.25-----> a2=2

the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}


N 4)

a²<span> + 5a − 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 5a+6.25) =14+6.25

Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}


N 5) 

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

(a-2.5)²=2025--------> (a-2.5)=(+/-)√20.25

a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

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A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding ma
IrinaVladis [17]

Answer:

538 books should be tested.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

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Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when M = 0.08

We have that \sigma = 0.72

M = z*\frac{\sigma}{\sqrt{n}}

0.08 = 2.575*\frac{0.72}{\sqrt{n}}

0.08\sqrt{n} = 2.575*0.72

\sqrt{n} = \frac{2.575*0.72}{0.08}

(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}

n = 537.1

Rounding up

538 books should be tested.

6 0
3 years ago
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