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3 years ago

How do you find missing values

1 answer:

5 0

Well, you could subtract or add or multiply or divide for example 45 + x=50 you could subtract 50 from 45 and get 5 there are many ways you can find missing values.

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Please help, only reply if you know answer

34kurt

Answer:

see explanation

Step-by-step explanation:

The translation represented by $\left[\begin{array}{ccc}1\\4\\\end{array}\right]$

interprets as a shift of 1 unit to the right ( add 1 to x- coordinate ) and a

shift of 4 units down ( subtract 4 from the y- coordinate ), then

(1, 4 ) → (1 + 1, 4 - 4 ) → (2, 0 )

(4, 4 ) → (4 + 1, 4 - 4 ) → (5, 0 )

(6, 2 ) → (6 + 1, 2 - 4 ) → (7, - 2 )

(1, 2 ) → (1 + 1, 2 - 4 ) → (2, - 2 )

8 0

3 years ago

Make y the subject: 5x + 2y = 11

Anton [14]

Does that mean solve for y? If so, here it is:
5x + 2y = 11.

Move the x term by subtracting.
2y = -5x + 11

Divide by 2.
y = -5/2x + 11/2

7 0

3 years ago

Read 2 more answers

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

(12a + 8) + (4a + 3) + (5 + a) =

PSYCHO15rus [73]

Answer:

17a + 16 =

Step-by-step explanation:

Add all a's

then all numbers w/o letters

you cannot add them together bc it would be like adding apples to carrots (thats what my teacher said)

3 0

2 years ago

SOMEONE PLS HELP :(!!!!!!

diamong [38]

It would be C as the answers

6 0

3 years ago