All categories

2 years ago

What is the domain for the function below? Х 1 3 5 7 Y 2 4 6 8

Mathematics

1 answer:

erma4kov [3.2K]2 years ago

3 0

X is the the answerrrr

You might be interested in

4x – 3y=8<br> 2. 5x – 2y = -11

AlladinOne [14]

Answer:

$x = -7$

$y = - 12$

Step-by-step explanation:

The given system is:

4x-3y=8

5x-2y=-11

We make x the subject in the top equation to get:

$x = \frac{3}{4}y + 2$

Plug this expression for x in the bottom equation to get

$5( \frac{3}{4}y + 2) - 2y= - 11$

We expand to obtain

$\frac{15}{4} y + 10 - 2y = - 11$

Multiply through by 4

$15y + 40 - 8y = - 44$

Group similar terms

$15y - 8y = - 44 - 40$

$7y = - 84$

$y = - \frac{84}{7} =-12$

This means

$x = -12\times \frac{3}{4} + 2$

$x = - 7$

6 0

3 years ago

Gretta is 1 1/2 meters tall. Which of the following is equivalent to 1 1/2 meters?

Furkat [3]

The answer would be one of the following in the photo below....if not do you have a picture you could show me of the possible answers?

7 0

2 years ago

Dave rolls a die and flips a coin. How many outcomes are in the sample space?

kogti [31]

Well the are six sides and numbers on the dices and two sides on the coin so there would be 8 outcomes

5 0

2 years ago

Read 2 more answers

From 12,000 graphing calculators produced by a manufacturer, an inspector selects a random sample of 550 calculators and finds

maria [59]

Answer:

131

Step-by-step explanation:

Using proportion,

If Out of 550 calculators = 6 are defective,

Then out of 12000 calculators = ? are defective

$\frac{12000 calculators}{550calculators} * 6$

= 21.8 * 6

= 130.8 ≈ 131

∴ Approximately 131 calculators are defective

3 0

2 years ago

Ab - bc+ ac, bc - ca+ ab, ca - ab-2bc

BabaBlast [244]

Answer:

$ab-2bc+ca$

Step-by-step explanation:

Complete question is given below

Add:-

ab-bc+ac,bc-ca+ab,ca-ab-2bc

We have to find the result after addition

Now, adding all expression

We get

$ab-bc+ac+bc-ca+ab+ca-ab-2bc$

Combine like terms then, we get

$(ab+ab-ab)+(bc-bc-2bc)+(ac-ca+ca)$

$ab-2bc+ca$

Hence, $ab-bc+ac+bc-ca+ab+ca-ab-2bc=ab-2bc+ca$

3 0

2 years ago