Can anyone just tell me how to do this please

Niech ktoś pomoże !<br> a)5 2\7 +1 3\7 -1\7=<br> b)4 5\9 -2 4\9 +3 2\9=<br> c)2 2\3 -(1 4\9 +2 3\9=

kvv77 [185]

A) 6 4/7

B) 5 3/9

C) 6 6/3

5 0

4 years ago

A car is traveling at a rate of 99 kilometers per hour. what is the car's rate in kilometers per minute? how many kilometers wil

gayaneshka [121]

there are 60 minutes per hour

99/60 = 1.65 km per minute

1.65 x 2 = 3.3 km in 2 minutes

7 0

4 years ago

The junior and senior students at Mathville High School are going to present an exciting musical entitled, "Math, What is it Goo

xeze [42]

Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

$\large\boxed{\large\boxed{161}}$

Explanation:

Let's put the information in a table step-by step.

(number of remaining students)

Juniors Seniors

Condition

Initially J S

15 seniors left S - 15

Twice juniors as seniors 2(S - 15)

3/4 of the juniors left 1/4×2(S - 15)
1/3 of seniors left 2/3×(S - 15)

At the end, there were 8 more seniors than juniors:

2/3×(S - 15) - 1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

$2/3\times (S - 15) - 1/4\times 2(S - 15) = 8$

Mutilply all by 12:

$8(S - 15)-6(S - 15)=96$

Distribution property:

$8S-120-6S-90=96$

Addtion property of equalities:

$8S-6S=96+120+90$

Add like terms:

$2S=306$

Division property of equalities:

$S=306/2=153$

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

$2/3\times (S-15)+1/4\times 2(S-15)$

Just substitute S with 153 fo find the number of students that remained to perfom in the musical:

$2/3\times (153-15)+1/4\times 2(153-15)\\ \\ 2/3(138)+1/2(138)$

$161$

5 0

3 years ago

What are the domain and range of f (x) = log (x minus 1) 2?.

statuscvo [17]

You can use the definition of logarithm and the fact that a positive number raised to any power will always stay bigger than 0.

The domain of the given function is {x | x > 1 and a real number }

The range of the given function is $\mathbb R$ (set of real numbers)

<h3>What is the definition of logarithm?</h3>

If a is raised to power b is resulted as c, then we can rewrite it that b equals to the logarithm of c with base a.

Or, symbolically:

$a^b = c \implies b = log_a(c)$

Since c was the result of a raised to power b, thus, if a was a positive number, then a raised to any power won't go less or equal to zero, thus making c > 0

<h3>How to use this definition to find the domain and range of given function?</h3>

Since log(x-1) is with base 10 (when base of log isn't specified, it is assumed to be with base 10) (when log is written ln, it is log with base e =2.71828.... ) thus, we have a = 10 > 0 thus the input x-1 > 0 too.

Or we have:

x > 1 as the restriction.

Thus domain of the given function is {x | x > 1 and a real number }

Now from domain, we have:

$x > 1\\
x-1 > 0\\
log(x-1) > -\infty\\
log(x-1) + 2 > -\infty\\
f(x) > -\infty$ (log(x-1) > -infinity since log(0) on right side have arbitrary negatively large value which is denoted by -infinity)