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3 years ago

Can someone just take an educated guess? Its multiple-choice and I'm so lost :(

Mathematics

2 answers:

Shkiper50 [21]3 years ago

6 0

Answer:

D is the right one im 100% sure on it

Step-by-step explanation:

White raven [17]3 years ago

6 0

Answer:

B or C

;-;

usually when I guess i usually go with c

I apologize if you get it wrong

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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

NEED HELP ASAP!!!!!!!!!!! 2. Solve for x 3. Solve for x

Anton [14]

Answer:

2. x = 47

3. x = 2

Step-by-step explanation:

These problems involve proportions, or equivalent ratios. You can solve for 'x' in each by using cross-multiplication and division.

2. 28(7) = 4(x + 2)

Distribute = 196 = 4x + 8

Subtract 8 from both sides: 196 - 8 = 4x + 8 - 8 or 188 = 4x

Solve for x: x = 47

3. 2(2x + 7) = 11(3x - 4)

Distribute: 4x + 14 = 33x - 44

Add 44 to both sides: 4x + 14 + 44 = 33x - 44 + 44 or 4x + 58 = 33x

Subtract 4x from both sides: 4x + 58 - 4x = 33x - 4x or 58 = 29x

Solve for x: x = 2

4 0

3 years ago

A box is to be made where the material for the sides and the lid cost $0.20 per square foot and the cost for the bottom is $0.

Reika [66]

Answer:

The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³

Step-by-step explanation:

From the given information;

Let a be the cost of the box

Let b be one side of the square base ; &

h to be the height of the box

We know that the volume of the box = 10 cubic feet

Then;

a²h = 10

h = 10/a²

The base = (0.65)a²

The top = (0.2)a²

The side = (0.2) a × 25/a²

= 5/a

For the four sides of the box now ;

= (0.2) 4a × 25/a²

= 0.8 × 25/a

= 20 /a

The total cost of the box is:

b = 0.65a² + 0.2a² + 20 /a

b = 0.85 a² + 20 /a

Taking differential of b with respect to a ;we have:

db/da = 1.7a - 1/a²(20) = 0

1.7 a³ - 20 = 0

1.7 a³ = 20

a³ = 20/1.7

a³ = 11.77

a = $\sqrt[3]{11.77}$

a = 2.274 ft

Thus; the cost for the base of the box = (0.65)a²

the cost for the base of the box =(0.65) × ( 2.274)²

the cost for the base of the box = 3.362

The top of the box = (0.2)a²

The top of the box = (0.2)× ( 2.274)²

The top of the box = 1.034

The four sides of the box = 20 /a

The four sides of the box = 20/2.274

The four sides of the box = 8.795

the total cost = b = 0.85 a² + 20 /a

the total cost = 0.85 (2.274)² + 20 /2.274

the total cost = 4.395 + 8.795

the total cost = 13.19

Recall that:

the volume of the box = 10 cubic feet

Then;

a²h = 10

h = 10/a²

h = 10/ 2.274²

h 1.934

The dimension of the box is l×w×h = (2.274 × 2.274 × 1.934) ft³

7 0

3 years ago

Please help me, I have minimum time

natita [175]

X= 5x Y=-6 I just added them together

8 0

2 years ago

What THREE integers have a sum of 300?

quester [9]

Answer:

Here we will use algebra to find three consecutive integers whose sum is 300. We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 300. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 300

To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 300

3X + 3 = 300

3X + 3 - 3 = 300 - 3

3X = 297

3X/3 = 297/3

X = 99

Which means that the first number is 99, the second number is 99 + 1 and the third number is 99 + 2. Therefore, three consecutive integers that add up to 300 are 99, 100, and 101.

99 + 100 + 101 = 300

We know our answer is correct because 99 + 100 + 101 equals 300 as displayed above.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers