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3 years ago

Using substitution for systems of equations, solve, please show work. thank you!!

Mathematics

2 answers:

Rasek [7]3 years ago

5 0

That's your final answer. X=7 and Y=2

ratelena [41]3 years ago

4 0

X is equal to 7 and Y is equal to 2

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Suppose x = 5 is a solution to the equation 4x − 3(x + a) = 2. Find the value of a that makes the equation true. Group of answer

xz_007 [3.2K]

Just replace all of the x’s with 5 and solve using order of operations. I will post what I got for my answer in the comments of the answer if you need it.

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3 years ago

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Virty [35]

Answer:

5 inches per hour

Step-by-step explanation:

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3 years ago

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A car with 15 gallons of gasoline drives 420 miles until the tank is empty. Write a linear equation that describes the amount of

storchak [24]

Answer:

$y = -\dfrac{1}{28}x + 15$

Step-by-step explanation:

The car uses 15 gallons of gasoline to drive 420 miles.

(15 gal)/(420 mi) = 1/28 gal/mi

The car uses 1/28 gallon of gasoline per mile.

y = mx + b

When x = 0, at the start of the drive, the car has 15 gallons of gasoline.

y = mx + 15

Then for every mile it travels, the amount of gasoline goes down by 1/28 gal. For x miles of travel, it uses 1/28 * x gallons of gasoline.

$y = -\dfrac{1}{28}x + 15$

5 0

3 years ago

Isaac records the following temperatures (in degrees Fahrenheit) at noon during one week:

Alecsey [184]

Answer:

first, we put them in order

78,[80],83,(85),86,[87],90

Q1 = 80

Q2 (median) = 85

Q3 = 87

Interquartile range (IQR) = Q3 - Q1 = 87 - 80 = 7 <==

Hope this helps!

5 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

4 years ago