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3 years ago

Answer correctly please !!!!! Will mark Brianliest !!!!!!!!!!!!!!

Mathematics

1 answer:

Paraphin [41]3 years ago

4 0

Answer:

7.18

Step-by-step explanation

4* COS(23) = 3.682

3.5+3.682 = 7.182

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Can someone please solve this?

forsale [732]

$|-8-6|=|-14|=14$

4 0

3 years ago

PLS HELP Choose the solution to this inequality. 43<−83y y < −12 y > 12 y < 43 y > 4

OlgaM077 [116]

Given:

The inequality is:

$\dfrac{4}{3}$

To find:

The solution of the given inequality.

Solution:

We have,

$\dfrac{4}{3}$

Multiply both sides by 3.

$4$

Divide both sides by -8 and change the sign of inequality.

$\dfrac{4}{-8}>\dfrac{-8y}{-8}$

$-\dfrac{1}{2}>y$

It can be written as:

$y$

Therefore, the correct option is A.

8 0

2 years ago

Drag each tile to the correct box. Not all tiles will be used.

Alina [70]

Answer:

3rd box, last box, 2nd box, 5th box in that order

Step-by-step explanation:

PEMDAS

7 0

3 years ago

Read 2 more answers

What are the coordinates of the point on the directed line segment from (-4,1) to (1,6) that partitions the segment into a ratio

Aleksandr [31]

Answer:

a segment is partitioned at a ratio of 1:3, then the point is one-fourth of the distance from (-4,-1) to (2,7).

To compute the x-coordinate of that point, you will need to compute one-fourth of the x-distance between 2 and -4 then add it to -4: (2--4)/4 = 1.5; 1.5 + -4 = -2.5.

To compute the y-coordinate of that point, you will need to compute one-fourth of the y-distance between 7 and -1 then add it to -1: (7--1)/4 = 2; 2 + -1 = 1.

The point is (-2.5,1)

3 0

2 years ago

Solve the following equation. X cubed minus 6X squared plus 6X equals zero

anyanavicka [17]

We have to solve this equation:

$x^3-6x^2+6x=0$

Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

$\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}$

Now we can find the roots of the quadratic polynomial as:

$\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}$

Then, the solutions to the equation are:

x = 0

x = 3 - √3

x = 3 + √3

4 0

1 year ago