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2 years ago

Can someone explain to me how to do this problem???

Mathematics

1 answer:

balandron [24]2 years ago

3 0

Answer is C. Mean is the mid-point value and one standard deviation on either side. (100-85=15 and 115-100=15)

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Andreus made $625 mowing yards. He worked for 5 consecutive days and earned the same amount of money each day. How much money di

olasank [31]

625 / 5 = 125 per day

4 0

3 years ago

What is the equation of the line that is parallel to the given

finlep [7]

Answer

O 4x+3y=-6

Step-by-step explanation:

Hello, I can help you witht this,

to solve this you need

step one

the line we are looking for is paraller to the line in the graph, it means both of then have the same slope.

find the slope of the line in the graph

remember

to find the slope of a line with two known points you can use

Let

$P1(x_{1},y_{1})}\\P1(x_{2},y_{2})\\slope\\s=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

pick two points from he graph

P1(0,3)

P2(3,-1)

put the values into the equation

$slope(m)\\m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\m=\frac{-1-3}{3-0}\\m= \frac{-4}{3}$

so, the slope of the line we are looking for is m=-4/3 and passes through the point (-3,2)

step 2

use the equation

$y-y_{1}=m(x-x_{1})$

to ding the equation of the line

P1(-3,2)

$y-y_{1}=m(x-x_{1})\\y-2=\frac{-4}{3} (x+3})\\isolate\ y\\y-2=\frac{-4x}{3}-4\\y=\frac{-4x}{3}-4+2\\y+\frac{4x}{3}=-2\\multiply\ each\ side\ by\ 3\\3(y+\frac{4x}{3})=3*-2\\\\3y+4x=-6\\4x+3y=-6\\$

and that's all, that is the equation of the line that is parallel to the given

line and passes through the point (-3, 2).

Have a good day.

7 0

3 years ago

Find the measure of angle X in the triangle below.

Elan Coil [88]

Answer:

X = 89.92° or 90.08°

Step-by-step explanation:

The law of sines can be used to find the value of angle X:

sin(X)/26 = sin(67.38°)/24

sin(X) = (26/24)sin(67.38°) ≈ 0.99999901787

There are two values of X that have this sine:

X = arcsin(0.99999901787) ≈ 89.92°

X = 180° -arcsin(0.99999901787) ≈ 90.08°

There are two solutions: X = 89.92° or 90.08°.

_____

<em>Comment on the problem</em>

We suspect that the angle is supposed to be considered to be 90°. However, the given angle is reported to 2 decimal places, so we figure the requested angle should also be reported to 2 decimal places.

The lengths of the short side that correspond to the above angles are 10.03 and 9.97 units. If the short side were considered to be 10 units, the triangle would be a right triangle, and the larger acute angle would be ...

arcsin(24/26) ≈ 67.38014° . . . . rounds to 67.38°

This points up the difficulty of trying to use the Law of Sines on a triangle that is actually a right triangle.

3 0

2 years ago

Read 2 more answers

I really need it to be sold in imaginary numbers

Yuliya22 [10]

Solving a 5th grade polynomial

We want to find the answer of the following polynomial:

$x^5+3x^4+3x^3+19x^2-54x-72=0$

We can see that the last term is -72

We want to find all the possible numbers that can divide it. Those are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

We want to factor this polynomial in order to find all the possible x-values. In order to factor it we will have to find some binomials that can divide it using the set of divisors of -72.

We know that if

(x - z) is a divisor of this polynomial then z might be a divisor of the last term -72.

We will verify which is a divisor using synthetic division. If it is a divisor then we can factor using it:

Let's begin with

(x-z) = (x - 1)

We want to divide

$\frac{(x^5+3x^4+3x^3+19x^2-54x-72)}{x-1}$

Using synthetic division we have that if the remainder is 0 it will be a factor

We can find the remainder by replacing x = z in the polynomial, when it is divided by (x - z). It is to say, that if we want to know if (x -1) is a factor of the polynomial we just need to replace x by 1, and see the result:

If the result is 0 it is a factor

If it is different to 0 it is not a factor

Replacing x = 1

If we replace x = 1, we will have that:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ 1^5+3\cdot1^4+3\cdot1^3+19\cdot1^2-54\cdot1-72 \\ =1+3+3+19-54-72 \\ =-100 \end{gathered}$

Then the remainder is not 0, then (x - 1) is not a factor.

Similarly we are going to apply this until we find factors:

(x - z) = (x + 1)

We replace x by -1:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ \downarrow \\ (-1)^5+3\cdot(-1)^4+3\cdot(-1)^3+19\cdot(-1)^2-54\cdot(-1)-72 \\ =-1+3-3+19+54-72 \\ =0 \end{gathered}$

Then, (x + 1) is a factor.

Using synthetic division we have that:

Then:

$x^5+3x^4+3x^3+19x^2-54x-72=(x+1)(x^4+2x^3+x^2+18x-72)$

Now, we want to factor the 4th grade polynomial.

Let's remember our possibilities:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

Since we verified ±1, let's try with ±2 as we did before.

(x - z) = (x - 2)

We want to divide:

$\frac{x^4+2x^3+x^2+18x-72}{x-2}$

We replace x by z = 2:

$\begin{gathered} x^4+2x^3+x^2+18x-72 \\ \downarrow \\ 2^4+2\cdot2^3+2^2+18\cdot2-72 \\ =16+16+4+36-72 \\ =0 \end{gathered}$

Then (x - 2) is a factor. Let's do the synthetic division:

Then,

$x^4+2x^3+x^2+18x-72=(x-2)(x^3+4x^2+9x+36)$

Then, our original polynomial is:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =\mleft(x+1\mright)\mleft(x^4+2x^3+x^2+18x-72\mright) \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \end{gathered}$

Now, let's prove if (x +2) is a factor, using the new 3th grade polynomial.

(x - z) = (x + 2)

We replace x by z = -2:

$\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-2)^3+4(-2)^2+9(-2)+36 \\ =-8+16-18+36 \\ =26 \end{gathered}$

Since the remainder is not 0, (x +2) is not a factor.

All the possible cases are:

{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}

let's prove with +4

(x - z) = (x + 4)

We want to divide:

$\frac{x^3+4x^2+9x+36}{x+4}$

Let's replace x by z = -4 in order to find the remainder:

$\begin{gathered} x^3+4x^2+9x+36 \\ \downarrow \\ (-4)^3+4(-4)^2+9(-4)+36 \\ =-64+64-36+36 \\ =0 \end{gathered}$

Then (x + 4) is a factor. Let's do the synthetic division:

Then,

$x^3+4x^2+9x+36=(x+4)(x^2+9)$

Since

x² + 9 cannot be factor, we have completed our factoring:

$\begin{gathered} x^5+3x^4+3x^3+19x^2-54x-72 \\ =(x-1)(x-2)(x^3+4x^2+9x+36) \\ =(x-1)(x-2)(x+4)(x^2+9) \end{gathered}$

Now, we have the following expression:

$(x-1)(x-2)(x+4)(x^2+9)=0$

Then, we have five posibilities:

(x - 1) = 0

or (x - 2) = 0

or (x + 4) = 0

or (x² + 9) = 0

Then, we have five solutions;

x - 1 = 0 → x₁ = 1

x - 2 = 0 → x₂ = 2

x + 4 = 0 → x₃ = -4

x² + 9 = 0 → x² = -9 → x = ±√-9 = ±√9√-1 = ±3i

→ x₄ = 3i

→ x₅ = -3i

<h2><em>Answer- the solutions of the polynomial are: x₁ = 1, x₂ = 2, x₃ = -4, x₄ = 3i and x₅ = -3i</em></h2>

7 0

10 months ago

ANSWER QUICK <br> 13+5 3/4

Vaselesa [24]

Answer:

18.75 or 18 3/4

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers