Find the length of side x in simplest radical form with a rational denominator.
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Could I get some help with this

Answer:

33/14

Step-by-step explanation:

[tex] - 10 \frac{2}{7} + ( - 4 \frac{4}{11} )[/tex]

[tex] = - \frac{72}{7} \div - ( \frac{48}{11} )[/tex]

[tex] = \frac{72}{7} \times \frac{11}{48} [/tex]

[tex] = \frac{3}{7} \times \frac{11}{2} [/tex]

[tex] = \frac{33}{14} [/tex]

stan dreamcatcher

Answer:

1≤f(x)≤5

Step-by-step explanation:

-1≤x≤1

-2≤2x≤2 (Multiplied by 2 both side)

-2+3≤2x+3≤2+3 (Adding three both sides)

1≤f(x)≤5

Answer:

k = 14

Step-by-step explanation:

Prime factorize 32

32 = 2 * 2 * 2 * 2 * 2  = 2⁵

[tex]\frac{2^{12}}{2^{\frac{k}{2}}}= 32\\\\\frac{2^{12}}{2^{\frac{k}{2}}}=2^{5}\\\\2^{12-\frac{k}{2}}=2^{5}[/tex]

Both sides base are same.So, compare exponents

[tex]12-\frac{k}{2}=5\\[/tex]

Subtract 12 from both side

[tex]-\frac{k}{2}=5-12\\\\-\frac{k}{2}=-7\\[/tex]

Multiply both sides by (-2),

[tex](-2)*(-\frac{k}{2})=-7*(-2)\\\\k = 14[/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:

{1, 0, 16}

Step-by-step explanation:

given..

j(x) = x^2-2x+1

put all given values of domain (1,0 and 5 ) in the equation..

the values you get are range of the function

Answer: 1

Step-by-step explanation

0(2)-2(0)+1=0+0+1= 1

1(1)-(2)(1)+1=1-2+1= 1

(5)(2)-(2)(5)+1=10-10+1= 1

Answer:

-117 is irrational number

Answer:

Irrational

Step-by-step explanation:

Irrational number can't be written as a faction, -11pie can't be written as a fraction. Therefore it is a irrational number.

[tex]\\ \sf\longmapsto 2x^2+3x^2+3(-1)=5x.x+5x.1[/tex]

[tex]\\ \sf\longmapsto 5x^2-3=5x^2+5x[/tex]

[tex]\\ \sf\longmapsto 5x^2-5x^2-3=5x[/tex]

[tex]\\ \sf\longmapsto 5x=-3[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{-3}{5}[/tex]

Answer:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

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.

the answer is

A. 28, 25, 22

[tex]\\ \sf\longmapsto Interest=Rate\times Time[/tex]

[tex]\\ \sf\longmapsto Interest=5.3(10)[/tex]

[tex]\\ \sf\longmapsto Interest =53[/tex]

It means that if your principal is $100 after 10years you will receive $53 as interest

Step-by-step explanation:

[tex]thank \: you[/tex]

Answer:

1950

Step-by-step explanation:

because 1950 column has the most highest number which is 11

Answer:

Between 1945 and 1950 ( from 5 to 11 ),

Step-by-step explanation:

Answer:

6.8 gallions i believe. im not quite sure

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]

Answer:

12

Step-by-step explanation:

for this you need to draw a straight line and plot the points J,K,L,M,N according to the question.

here,

given K lies between J and M, L is between K and M , M is between K and N.

JN= 30, KM=8

JK=KL=LM

MJ= ?

according to the question,

J--------K-------L--------M-------N

KM= 8

KL+LM= 8 (KL+LM=KM)

KL+KL=8 ( given, LM=KL)

2KL= 8

KL=4

so,

JK= KL=4 and LM= KL= 4

now,

MJ= ?

we know

MJ= JK+KM = 4+8= 12

Answer:

Let ABC = 73.6

Complement = ABD = 16.4

ABx = unknown angle

ABx + (ABx + 73.6) = 90

ABx = 16.4 / 2 = 8.2

The angles are 8.2 and (8.2 + 73.6) = 90

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.