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2 years ago

|x| + |y| Evaluate the expression if x = 16 and y = 13.

Mathematics

2 answers:

Schach [20]2 years ago

8 0

Answer:

|x| + |y| = 29

Step-by-step explanation:

We have been given the expression;

|x| + |y|

We are required to evaluate the above expression if;

x = 16 and y = 13

The parallel vertical lines imply that we are dealing with the absolute value of x and y;

|x| + |y| = |16| + |13|

= 16 + 13

= 29

Ede4ka [16]2 years ago

4 0

Answer:

The correct answer is 29

Step-by-step explanation:

<u>Points to remember</u>

|x| = x

|-x| = x

It is given an expression, |x| + |y|

<u>To find the value of given expression</u>

|x| + |y| if x = 16 and y = 13

Substitute the value of x and y in the above expression,

|x| + |y| = |16| + |13|

= 16 + 13 = 29

Therefore the correct answer is 29

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1 tenth is how many times greater than 1 hundredth

Anna35 [415]

Answer:

Step-by-step explanation:

1 tenth: 1/10

1 hundredth: 1/100

1/10 = X × 1/100

X = 100/10

X = 10

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3 years ago

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A ladder is placed against a wall, such that it touches the top of the wall 6 meters above the ground. The ladder is inclined at

luda_lava [24]

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2 years ago

Marty made a $220 bank deposit using $10 bills and $5 bills. She gave the teller a total of 38 bills, how many $5 bills were in

Luda [366]

ANSWER: 32 five-dollar bills

======

EXPLANATION:

Let x be number of $5 bills

Let y be number of $10 bills

Since we have total of 38 bills, we must have the sum of x and y be 38

x + y = 38 (I)

Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):

5x + 10y = 220 (II)

System of equations:

$\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ 5x + 10y &= 220 && \text{(II)} \end{aligned} \right.$

Divide both sides of equation (II) by 5 so our numbers become smaller

$\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ x + 2y &= 44 && \text{(II)} \end{aligned} \right.$

Rearrange (I) to solve for y so that we can substitute into (II)

$\begin{aligned} x + y &= 38 && \text{(I)} \\ y &= 38 - x \end{aligned}$

Substituting this into equation (II) for the y:

$\begin{aligned} x + 2y &= 44 && \text{(II)} \\ x + 2(38 - x) &= 44\\ x + 76 - 2x &= 44 \\ -x &= -32 \\ x &= 32 \end{aligned}$

We have 32 five-dollar bills

======

If we want to finish off the question, use y = 38 - x to figure out number of $10 bills

$y = 38 - 32 = 6$

32 five-dollar bills and 6 ten-dollar bills

7 0

3 years ago

Can someone plz explain how to do this

swat32

Answer:

-2

Step-by-step explanation:

First, you need to find the equation.

f(x) = -3x + b

Now we need to find the y-intercept.

f(x) = -3x + b

f(-9) = -3(1) + b

-9 = -3 + b

-6 = b

f(x) = -3x - 6

The zero of f means that f(x) = 0

f(0) = -3x - 6

f(6) = -3x

x = -2

5 0

2 years ago

If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.

Lelechka [254]

Answer:

$\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }$

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

$f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)$

So we can say that

$a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}$

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

$(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab$

and we can notice that

$a^2+b^2=(a+b)^2-2ab$ so

$(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab$

it comes

$x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}$

multiply by 3

$3x^2-20x+2*16+5=3x^2-20x+37$

4 0

2 years ago