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3 years ago

B. 2(3x - 5) = 2x + 6

Mathematics

2 answers:

Leto [7]3 years ago

8 0

Answer:

x = 4

Step-by-step explanation:

Given

2(3x - 5) = 2x + 6 ( divide both sides by 2 )

3x - 5 = x + 3 ( subtract x from both sides )

2x - 5 = 3 ( add 5 to both sides )

2x = 8 ( divide both sides by 2 )

x = 4

gogolik [260]3 years ago

6 0

Answer:

x = 4

Step-by-step explanation:

2(3x - 5) = 2x + 6

expand- 2(3x-5):6x+10

6x-10= 2x+6

Add 10 to both sides

6x-10+10=2x+6+10

simplify

6x=2x+16

subtract 2x from both sides

6x-2x=2x+16-2x

simplify

4x=16

divide both sides by 4

4x/4=16/4

x=4

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MATH PLEASE HELP 50 POINTS 1 QUESTION

miv72 [106K]

For 1, ∠1 and ∠5 are supplementary is a given, so you can just write 'Given'
For 2, due to that 1 and 2 are vertical angles (angles that are opposite to each other when two lines cross according to Math is Fun)
For 3, since ∠1=∠2, ∠1 can be substituted by ∠2 in ∠1+∠5=180 to get ∠2+∠5=180. Therefore, they are supplementary.

For 4, due to that we have Consecutive Interior Angles (they are angles that are on one side of a transversal, which is the line that cuts through a and b,but inside the two lines according to Math is Fun) adding up to 180 degrees, a || b

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3 years ago

Which number is not irrational? A. 20 B. 127 c. 125 D. V50

Tju [1.3M]

Answer:

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3 years ago

Write a comparison sentence that includes the words as long as and compares

Paha777 [63]

Jenny pencil is as long as a giraffe's neck.

5 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:

18 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 18 values for n are possible.

3 0

3 years ago

On Heiki's trip, the ratio of distance traveled to time is 112 km to 3 h.

allsm [11]

336

112x = 336

112
x3
=336

37.3
x3
= 112

37.3
x9
=336

6 0

3 years ago