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3 years ago

23=x-9;x=14 how do I figure out if this is a solution of the equation?

2 answers:

5 0

Subsitute 14 back in for x. So you should get
23 = (14) - 9
If this is correct, both sides should equal each other.

However, it is not, because 23 $\neq$ 5.

x=32 is the solution.

serious [3.7K]3 years ago

4 0

There is no solution because x is incorrect.

x should be 32

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I need help please answer these two questions for me! Will give points.

jarptica [38.1K]

Answer:

7.) 7

10.) 0

Step-by-step explanation:

When it means "evaluate the function", it's in essence asking us to see what the function spits out when we feed it a certain input. Our inputs are our x values, which spit out a y value.

Evaluating the function when x = 1:

Let's look at where the function has an x value of 1. We see it near the bottom of the table and see the y value associated with the input is 7. So when the function is fed 1 as an input, it spits out 7.

Evaluating the function when f(x) = - 2:

This one is a weird because of the new notation. Just think of it as some value of f, which we don't know (so we represent it as an x-variable) must equal -2. So let's look at our table to find out where our output is -2. We find that when f(x) = -2 the input is 0. So the input which gives -2 is 0.

7 0

3 years ago

Find the solution to the equation z – 7 = 8.

Bas_tet [7]

Z - 7 = 8

subtract 7 from both sides

z = 15

hope this helped!

7 0

2 years ago

Read 2 more answers

In a bag of sweets, 6 are orange and 9 are lime. Jodi chooses a sweet at random, eats it an

Digiron [165]

Answer: $\frac{17}{35}$

Step-by-step explanation:

Given

The bag contains 6 orange and 9 lime sweets

For both sweets of the same flavor, Jodi must either choose orange or lime

No. of ways it can be done is $^6C_1\times^5C_1+^9C_1\times^8C_1$

The total no of ways of selecting two sweets out of 15 is $^{15}C_1\times^{14}C_1$

The probability that both sweets are the same flavor

$P=\dfrac{^6C_1\times^5C_1+^9C_1\times^8C_1}{^{15}C_1\times^{14}C_1}=\frac{102}{210}=\frac{51}{105}=\frac{17}{35}$

3 0

3 years ago

How many different 10-letter words (real or imaginary) can be formed from the letters of the word LITERATURE?

nordsb [41]

Answer:

The number of words that can be formed from the word "LITERATURE" is 453600

Step-by-step explanation:

Given

Word: LITERATURE

Required: Number of 10 letter word that can be formed

The number of letters in the word "LITERATURE" is 10

But some letters are repeated; These letters are T, E and R.

Each of the letters are repeated twice (2 times)

i.e.

Number of T = 2

Number of E = 2

Number of R = 2

To calculate the number of words that can be formed, the total number of possible arrangements will be divided by arrangement of each repeated character. This is done as follows;

Number of words that can be formed = $\frac{10!}{2!2!2!}$