All categories

3 years ago

–0.4(3x – 2) + StartFraction 2 x plus 4 Over 3 EndFraction for x = 4

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

Given

- 0.4(3x - 2) + $\frac{2x+4}{3}$ ← substitute x = 4 into the expression

= - 0.4(3(4) - 2) + $\frac{2(4)+4}{3}$

= - 0.4(12 - 2) + $\frac{8+4}{3}$

= - 0.4(10) + $\frac{12}{3}$

= - 4 + 4 = 0

kenny6666 [7]3 years ago

3 0

0 also thanks for putting the whole thing most people don’t so it hard to understand lol

You might be interested in

In an election, a candidate got 82% votes. He won the beating his rival candidate by 576 votes. How many votes were polled?

Slav-nsk [51]

Answer:

900

Step-by-step explanation:

A candidate got 82%votes.

Loosing candidate get=(100-82)=18%

Difference between two votes.

=(82-18)%=64%

Let be the Total vote is=x.

64% of the total vote=576

$Or, 64 \%\times x= 576$

$Or, \frac{64}{100} \times x = 576$

$Or,x= \frac{576 \times 100}{64}$

$Or,x=900$

∴The total votes that were polled are 900.

8 0

3 years ago

Maya has 120 caramel apples to sell. each caramel apple is covered with one topping

oksano4ka [1.4K]

Answer:

1/5 * 120 = 24 peanut

1/3 * 120 = 40 chocolate chips

3/10 * 120 = 36 coconut

There are 20 left

Unless there are other toppings not mentioned,

those last 20 must have sprinkles

4 0

3 years ago

Suppose the roots of the polynomial $x^2 - mx + n$ are positive prime integers (not necessarily distinct). Given that $m < 20

Vsevolod [243]

Answer:

<em>18</em> values for n are possible.

Step-by-step explanation:

Given the quadratic polynomial:

$x^2 - mx + n$

such that:

Roots are positive prime integers and

$m < 20$

To find:

How many possible values of $n$ are there ?

Solution:

First of all, let us have a look at the sum and product of a quadratic equation.

If the quadratic equation is:

$Ax^{2} +Bx+C$

and the roots are: $\alpha$ and $\beta$

Then sum of roots, $\alpha+\beta = -\frac{B}{A}$

Product of roots, $\alpha \beta = \frac{C}{A}$

Comparing the given equation with standard equation, we get:

A = 1, B = -m and C = n

Sum of roots, $\alpha+\beta = -\frac{-m}{1} = m$

Product of roots, $\alpha \beta = \frac{n}{1} = n$

We are given that $m$

$\alpha$ and $\beta$ are positive prime integers such that their sum is less than 20.

Let us have a look at some of the positive prime integers:

2, 3, 5, 7, 11, 13, 17, 23, 29, .....

Now, we have to choose two such prime integers from above list such that their sum is less than 20 and the roots can be repetitive as well.

So, possible combinations and possible value of $n (= \alpha \times \beta)$ are:

$1.\ 2, 2\Rightarrow n = 2\times 2 = 4\\2.\ 2, 3 \Rightarrow n = 6\\3.\ 2, 5 \Rightarrow n = 10\\4.\ 2, 7\Rightarrow n = 14\\5.\ 2, 11 \Rightarrow n = 22\\6.\ 2, 13 \Rightarrow n = 26\\7.\ 2, 17 \Rightarrow n = 34\\8.\ 3, 3\Rightarrow n = 3\times 3 = 9\\9.\ 3, 5 \Rightarrow n = 15\\10.\ 3, 7 \Rightarrow n = 21\\$

$11.\ 3, 11\Rightarrow n = 33\\12.\ 3, 13 \Rightarrow n = 39\\13.\ 5, 5 \Rightarrow n = 25\\14.\ 5, 7 \Rightarrow n = 35\\15.\ 5, 11 \Rightarrow n = 55\\16.\ 5, 13 \Rightarrow n = 65\\17.\ 7, 7 \Rightarrow n = 49\\18.\ 7, 11 \Rightarrow n = 77$

So,as shown above <em>18 values for n are possible.</em>

3 0

3 years ago

A ladder 15 feet from the base of a building reaches a window that is 35 feet high. What is the length of the ladder to the near

nikdorinn [45]

All you do is use the a+b=c formula lets plug it in
15+b=35
15+20=35
35=35
The answer is 20

7 0

3 years ago

Kris wanted to understand whether students at her school were in favor of an extended school day. She surveyed some students and

puteri [66]

the table below:

Grade In favor Opposed Undecided Total

Grade 9 6 4 8 18

Grade 10 10 11 9 30

Grade 11 12 15 11 38

Grade 12 15 6 14 35

If the principal randomly selects a student in grade 10 from this survey,

the probability that the student is opposed to extending the school day

P ( student in grade 10 is opposed)= Number of student in opposed(grade10) / total number of students in grade 10

Number of student opposed who is in grade10 = 11

total number of students in grade 10 = 30

P( student in grade 10 is opposed)= $\frac{11}{30}$

= 0.37

the probability that the student is opposed to extending the school day= 0.37

5 0

3 years ago