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3 years ago

Please help me it is due tonight

Mathematics

1 answer:

Aneli [31]3 years ago

8 0

Answer: with what

Step-by-step explanation:

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There are three apple orchards. Each orchard has a different number of trees and apples.

Assoli18 [71]

Answer:

A. The #1 Seed

Step-by-step explanation

To find the answer you find the rate of one apple tree to # of apples. you do this by dividing the # of apples from the orchard by the apple trees and whichever has the highest rate of Apples to Trees is the answer which would be The #1 Seed.

4 0

3 years ago

What is the domain and range of the following function

mel-nik [20]

Answer:

domain: (–∞,∞) or all real numbers

range: (–∞,∞) or all real numbers

Step-by-step explanation:

domain is left to right (x-axis). looking at the graph, as you go from left to right, the x-values are infinitely getting more negative and positive.

range is bottom to top (y-axis). as you go from bottom to top, the y-values are infinitely getting more negative and positive.

6 0

3 years ago

Using the figure below, select the pairs of complementary angles.

Mila [183]

Answer:

c) <DBE and <EBC

Step-by-step explanation:

This is because when added together, both angles equal 90 degrees.

Complementary angles always equal 90 degrees.

8 0

3 years ago

A rectangular sign has length of 3 and 1/2 ft and a width of 1 and 1/8 ft. Will the area of the sign be greater or less than 3 a

zavuch27 [327]

Answer:

$Area = 3\dfrac{15}{16}\ sq\ ft$

Area is greater than $3\frac{1}{2}$ sq ft.

Step-by-step explanation:

Length of rectangular sign board = $3\frac{1}{2}$ ft

Width of rectangular sign board = $1\frac{1}{8}$ ft

Area of a rectangle is given by the formula:

$A = Length \times Width$

To find the area, we are going to multiply $3\frac{1}{2}$ with $1\frac{1}{8}$ .

$1\frac{1}{8}$ is greater than 1.

Therefore the multiplication will be greater than $3\frac{1}{2}$ .

Hence, we can say that area will be greater than $3\frac{1}{2}$ sq ft.

Area of the sign:

$A = 3\dfrac{1}{2}\times 1\dfrac{1}8\\\Rightarrow A = \dfrac{7}{2}\times \frac{9}8\\\Rightarrow A = \dfrac{63}{16}\\\Rightarrow A = 3\dfrac{15}{16}\ sq\ ft$

8 0

3 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

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Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

3 years ago