Julie has 5.00 and wants to buy some glue at Walmart glue cost 4.25 how much will julie get back in change and you have to subtract a d slove problem

Answer:

-1,2 can be plotted together with these four points ...

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

Simplify,

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

[tex]2-\sqrt{3}=\frac{(\sqrt{3}-1)^2}{3-1}[/tex]

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer:

240

Step-by-step explanation:

We are given a right triangle. Based on the leg rule, the following equation shows how the length of a leg in a right triangle relates with the segments connected to the hypotenuse:

Hypotenuse/leg = leg/part

Where,

Hypo = 400

Leg = x

Part = 144

Substitute

400/x = x/144

Cross multiply

400*144 = x*x

57,600 = x²

√57,600 = x

240 = x

x = 240

Answer:

6.8 gallions i believe. im not quite sure

Answer:

(-1,-6)

Step-by-step explanation:

(13 + x)/2 = 6

13+x= 12

x = -1

~~~~~~~~~~~~~~~

(-2 + y ) / 2 = -4

-2 + y = -8

y = -6

The coordinates of the other endpoint will be (-1,-6). The correct option is C.

Divide the measurement of the distance between the two end locations by 2. The middle of that line is located at this separation from either end.

A coordinate plane is a 2D plane that is formed by the intersection of two perpendicular lines known as the x-axis and y-axis.

Given that the midpoint of a segment is (6,−4) and one endpoint is (13,−2).

The x- coordinate will be calculated as:-

(13 + x)/2 = 6

13+x= 12

x = -1

The y-coordinate will be calculated as:-

(-2 + y ) / 2 = -4

-2 + y = -8

y = -6

Therefore, the coordinates of the other endpoint will be (-1,-6). The correct option is C.

To know more about midpoints of the line follow

https://brainly.com/question/24431553

#SPJ2

Answer:

x = -0.5

Step-by-step explanation:

x + 0.25 = -0.25

x = -0.25 - 0.25

x = -0.5

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation.  Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}[/tex]

One is given the following equation,

[tex]\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0[/tex]

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

[tex]sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}[/tex]

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

[tex]\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0[/tex]

[tex]\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0[/tex]

[tex]\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}[/tex]

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

[tex]\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}[/tex]

[tex]\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}[/tex]

[tex]1+(-1)=0[/tex]

[tex]1-1=0[/tex]

[tex]0=0[/tex]

9514 1404 393

Answer:

  2. (only)

Step-by-step explanation:

The Pythagorean theorem tells you the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. To determine if these are right triangles, determine if that condition is met.

  1. 3^2 +5^2 = 9 + 25 = 34 ≠ (√35)^2 . . . . not a right triangle

  2. 5^2 +4^2 = 25 +16 = 41 = (√41)^2 . . . a right triangle

  3. 6^2 +8^2 = 36 +64 = 100 ≠ (√10)^2 . . . . not a right triangle

  4. 3^2 +3^2 = 9 +9 = 18 ≠ (3√3)^2 = 27 . . . . not a right triangle

Answer:

The area of the circle is 100 square units.

Step-by-step explanation:

We are given that the circumference of a circle is 20π, and we want to determine its area.

Recall that the circumference of a circle is given by the formula:

[tex]\displaystyle C = 2\pi r[/tex]

Substitute:

[tex]20 \pi = 2 \pi r[/tex]

Solve for the radius:

[tex]\displaystyle r = \frac{20\pi}{2\pi} = 10[/tex]

The area of a circle is given by:

[tex]\displaystyle A = \pi r^2[/tex]

Since the radius is 10 units:

[tex]\displaystyle A = \pi (10)^2[/tex]

Evaluate:

[tex]\displaystyle A = 100\pi\text{ units}^2[/tex]

In conclusion, the area of the circle is 100 square units.

Answer: [tex]w=\frac{33}{28}[/tex]

Step-by-step explanation:

To solve for w, we want to isolate w.

[tex]-\frac{9}{7}=-\frac{2}{3}w-\frac{1}{2}[/tex]            [add both sides by 1/2]

[tex]-\frac{11}{14}=-\frac{2}{3}w[/tex]                 [multiply both sides by -3/2]

[tex]w=\frac{33}{28}[/tex]

Now we know that [tex]w=\frac{33}{28}[/tex].

Answer:

[tex]\sf w=\dfrac{33}{28} \\[/tex]

Step-by-step explanation:

[tex]\sf -\dfrac{9}{7} =-\dfrac{2w}{3} -\dfrac{1}{2}[/tex]

First, take -2w/3 to the left side.

[tex]\sf -\dfrac{9}{7}+\dfrac{2w}{3} = -\dfrac{1}{2}[/tex]

Then, add 9/7 to both sides.

[tex]\sf \dfrac{2w}{3} = -\dfrac{1}{2}+\dfrac{9}{7}[/tex]

Make the denominators the same and add the fractions.

[tex]\sf \dfrac{2w}{3} = -\dfrac{1*7}{2*7}+\dfrac{9*2}{7*2}\\\\\sf \dfrac{2w}{3} = -\dfrac{7}{14}+\dfrac{18}{14}\\\\\sf \dfrac{2w}{3} = \dfrac{-7+18}{14}\\\\\sf \dfrac{2w}{3} = \dfrac{11}{14}[/tex]

Use cross multiplication.

[tex]\sf 2w*14=11*3\\\\28w=33[/tex]

Divide both sides by 28.

[tex]\sf w=\dfrac{33}{28} \\[/tex]

Step-by-step explanationIf you are reading this say

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