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3 years ago

Is (1, 4) a solution to this system of equations? 8x + 3y = 20 x + 4y = 17

Mathematics

1 answer:

vampirchik [111]3 years ago

6 0

Answer:

Yes.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)
Systems of Equations

Step-by-step explanation:

Step 1: Define

Identify

8x + 3y = 20

x + 4y = 17

(1, 4)

Step 2: Check

Plug and verify the coordinates is a solution.

Substitute in point [1st Equation]: 8(1) + 3(4) = 20
Multiply: 8 + 12 = 20
Add: 20 = 20 $\checkmark$
Substitute in point [2nd Equation]: 1 + 4(4) = 17
Multiply: 1 + 16 = 17
Add: 17 = 17 $\checkmark$

∴ Since both equations work out, (1, 4) is a solution to the systems of equations.

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Mary has $267 in her checking account. She writes checks for $33, $65, and $112. What is the balance in her account now?

nata0808 [166]

Answer:

267 - 112 - 65 - 33 = $57

Step-by-step explanation:

subtract to get the answer

5 0

3 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

Angelina_Jolie [31]

Answer:

$P(X = 0) = 0.027$

$P(X = 1) = 0.189$

$P(X = 2) = 0.441$

$P(X = 3)= 0.343$

Step-by-step explanation:

The probability mass function P(X = x) is the probability that X happens x times.

When n trials happen, for each $x \leq n$ , the probability mass function is given by:

$P(X = x) = pe_{n,x}.p^{x}.(1-p)^{n-x}$

In which p is the probability that the event happens.

$pe_{n,x},$ is the permutation of n elements with x repetitions(when there are multiple events happening(like one passes and two not passing)). It can be calculated by the following formula:

$pe_{n,x} = \frac{n!}{x!}$

The sum of all P(X=x) must be 1.

In this problem

We have 3 trials, so $n = 3$

The probability that a wafer pass a test is 0.7, so $p = 0.7$

Determine the probability mass function of the number of wafers from a lot that pass the test.

$P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027$

$P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189$

$P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441$

$P(X = 3) = (0.7)^{3} = 0.343$

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3 years ago

Tristan is working on dilating rectangles. Rectangle ABCD is dilated by a factor of 3. The resulting rectangle WXYZ has an area

zloy xaker [14]

Answer:

Tristan is correct, because he needs to find the possible dimensions for rectangle WXYZ and then divide each dimension by 3.

Step-by-step explanation:

The rectangle ABCD is dilated by a factor of 3 to get the rectangle WXYZ whose area is found to be 72 cm².

Let the dimensions of the dilated rectangle WXYZ are a cm by b cm.

So, ab = 72 ........... (1)

Now, the dimensions of the original rectangle are 3 times lesser than the dimensions of rectangle WXYZ.

So, the area of rectangle ABCD will be $(\frac{a}{3} \times \frac{b}{3}) = \frac{ab}{9} = \frac{72}{9} = 8$ cm² {from equation (1)}

Therefore, Tristan is correct, because he needs to find the possible dimensions for rectangle WXYZ and then divide each dimension by 3. (Answer)

8 0

2 years ago

Favian received a $100 gift card to a clothing store.

IgorLugansk [536]

Answer:

32n = 100 - 20

n = 2.5

he purchased 2 sweaters

Step-by-step explanation:

32n = 100 - 20

If Favian originally had $100 but spent $20 on a belt, he only had 80 left, meaning that 80 = 32n which means that he was able to but 2 sweaters and got $16 back. Therefore the formula would be 32n = 100 - 20

4 0

3 years ago

Factor out the GCF from the terms of the polynomial –4y5 + 6y3 + 8y2 – 2y. A. –2y4 + 3y2 + 4y – 1 B. y(–4y4 + 6y2 + 8y – 2) C. 2

andreyandreev [35.5K]

Answer:

Option D) $-2y(2y^4-3y^2-4y+1)$ is correct

Therefore $-4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)$

Step-by-step explanation:

Given polynomial is $-4y^5+6y^3+8y^2-2y$

To factorise the given polynomial by taking out the common terms of the given polynomial :

$-4y^5+6y^3+8y^2-2y$
$=2y(-2y^4+3y^2+4y-1)$ ( here 2y is GCF common term so taking outside the terms of the polynomial )
$=-2y(2y^4-3y^2-4y+1)$ ( now taking (-) outside )

Therefore $-4y^5+6y^3+8y^2-2y=-2y(2y^4-3y^2-4y+1)$

Option D) $-2y(2y^4-3y^2-4y+1)$ is correct

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3 years ago