solve for K
7K-6
2K+14

Answer:

3$

Step-by-step explanation:

if the sale is 50% you multiply the 6$ original price by the 50% discount to get 3$

Answer:

[tex](x,y) = (0,4)~ \text{and}~ (x,y) = (2,8)[/tex]

Step by step explanation:

[tex]y = x^2 +4~~~~~~~~~...(i)\\\\y = 2x +4~~~~~~~~~...(ii)\\\\\text{From (i) and (ii):}\\\\~~~~~~x^2 +4 = 2x+4\\\\\implies x^2 = 2x\\\\\implies x^2 -2x = 0\\\\\implies x(x-2) = 0\\\\\implies x =0,~ x = 2[/tex]

[tex]\text{Substitute x = 0 in eq (i):}\\\\y = 0^2 +4 = 4\\\\\text{Substitute x = 2 in eq (i):}\\\\y=2^2+4 = 4+4 = 8\\\\\text{Hence,}~ (x,y)= \{(2,8), (0,4)\}[/tex]

Jack and Jills' speed in miles per hour is 4

The given parameters are:

Jack speed = x^2-11x-22

Jill distance = x^2-3x-54 and time = x + 6

Speed is calculated as:

Speed = Distance/time

So, we have:

Jill speed = x^2-3x-54/x + 6

Factorize the numerator

Jill speed = (x + 6)(x - 9)/x + 6

Cancel the common factor

Jill speed = x - 9

Both speeds are equal.

So, we have:

x^2-11x-22 = x - 9

Collect like terms

x^2-11x - x -22 + 9 = 0

Evaluate

x^2 - 12x - 13 = 0

Factorize the expression

(x + 1)(x - 13) = 0

Solve for x

x = -1 or x = 13

x cannot be negative.

So, we have:

x =13

Substitute x =13 in Jill speed = x - 9

Jill speed = 13 - 9

Evaluate

Jill speed = 4

Hence, their speed in miles per hour is 4

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Answer:

[tex]\frac{100}{a}[/tex]

Step-by-step explanation:

Hello!

a% is the same as [tex]\frac a{100}[/tex].

a% of a can be found by multiplying them both, giving us [tex]\frac {a^2}{100}[/tex]

The value of the expression is [tex]\frac{100}{a}[/tex]

The additive inverse of the complex number –8 + 3i will be 8 – 3i. Then the correct option is C.

To produce an answer equal to 0, additive inverse simply entails switching the sign of the quantity and adding it to the original figure.

The complex expression is given below.

⇒ –8 + 3i

Then the additive inverse of the complex number will be

Let x be the additive inverse of the complex number.

Then we have

x + (–8 + 3i) = 0

x = –(–8 + 3i)

x = 8 – 3i

Then the correct option is C.

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Answer:

C

Step-by-step explanation:

Edge 2022

Answer:

x>11

Step-by-step explanation:

To simplify this inequality, we add 3 from both sides to get x-3+3>8+3. Since -3+3 cancels out to 0, we are left with x>8+3. 8+3=11 so x>11.

Hope this helps! If you need further explanation let me know :)

Answer:

First option.

Step-by-step explanation:

Look at the position of letters, you don't really need to look at the picture.

BQRT ~ XZGJ means figure BQRT is similar to figure XZGJ and congruent angles are written in same order.

So first letter of BQRT is B and first letter of XZGJ is X.

That means that angle B is congruent to angle X.

You can write that as: ∠B ≅ ∠X

Second letter of BQRT is Q and second letter of XZGJ is Z.

Angle Q is congruent to angle Z.

∠Q ≅ ∠Z

And same for last two pairs of angles.

Final answer:

∡B ≅ ∡X, ∡Q ≅ ∡Z, ∡R ≅ ∡G, ∡T ≅ ∡J

Answer:

Step-by-step explanation:

A = P(1-r)^n

A = 35098(1-2.33%)^13

   = $25 833.18

Answer:

A ≈ 300 cm²

Step-by-step explanation:

Area of a circle: πr² (where r is the radius)

Substitute the given values into the formula:

⇒ A = πr²

⇒ A ≈ (3)(10)²

⇒ A ≈ 3(100)

⇒ A ≈ 300 cm²

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Hey there!

[tex] \red{\underline{\bold{\blue{Area \: of \: a \: circle}}}} [/tex]

[tex] \\ [/tex]

[tex] \red{\boxed{\bold{\blue{A \approx 300{cm}^{2}}}}} [/tex]

The area of a circle is calculated using the following formula :

[tex] \sf{\blue{A} = \orange{\pi} \times \red{r}^{2}} \\ \sf{where \: \red{r}\: is \: the \: radius \: of \: the \: circle} [/tex]

⇢Here, we consider that π = 3 but in reality, π ≈ 3 replace [tex] \orange{\pi} [/tex] with 3 and [tex] \red{r} [/tex] with 10 :

[tex] \sf{\blue{A} = \orange{\pi} \times \red{r}^{2} } \: \: \: \\ \sf{\blue{A} \approx \orange{3} \times \red{10}^{2} } \\ \sf{\blue{A} \approx \orange{3} \times \red{100} } \\ \green{ \boxed{\sf{\blue{A} \approx300{cm}^{2}}}} [/tex]

⇒Therefore, the area of the circle is about 300cm².

[tex] \\ [/tex]

Answer:

1/3 or 33%

Step-by-step explanation:

The area of the parallelogram of height 9 in and base 13 in will be 117in².

The area of a parallelogram is the product of the base and the height of the figure.

The area of a parallelogram = Base x Height

WE have given a parallelogram of height 9 in and base 13 in.

So,

The area of a parallelogram = Base x Height

A = Base x Height

A = 13 × 9

A = 117in²

Hence, the area of the parallelogram will be 117in².

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Answer:

26cm

Step-by-step explanation:

Answer:

0

Step-by-step explanation:

0, it is a flat line on a graph

Answer:

A.

Step-by-step explanation:

It says he scored five points less than twice the number of points he scored last week.

2x-5=25

the 2x represents the number of points last week.

the -5 represents the five points less.

the number 25 is the points this week.

Hope this helps!

If not, I am sorry.

Answer:

Because Rhomboids have adjacent sides that makes it equal, and also both sides have the same meeting point which is M.

Step-by-step explanation:

The number that the digit 1 has a value that is ten times as great as the value of the digit 1 in the number 215,784 is 2.

Place value can be defined as the value that's represented by a digit on the basis of the position of the digit.

It should be noted that the value of the digit 1 in 215784 is 10000. On the other hand, the value of 2 is 200000. It's in hundred of thousands. Therefore, it's 10 × (10000).

Therefore the digit is 2.

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Answer:

it is 103,748

Step-by-step explanation:

because it was wrong when I pressed 15,839

Answer:

2√3

Step-by-step explanation:

The shown triangle is a 30-60-90 special right triangle due to the right angle and the thirty degree angle(the other angle has to be 60 degrees).

The hypotenuse of a 30-60-90 triangle is twice the leg opposite the 30 angle. The leg opposite the 30 degree angle is √3, so x would be 2√3.

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

Let's solve for x ~

We can apply Trigonometric ratios here, since the given triangle is a right angled triangle

so,

[tex]\qquad \sf  \dashrightarrow \: \sin(30) = \dfrac{ \sqrt{3} }{x} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{1}{2} = \dfrac{ \sqrt{3} }{x} [/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 2 \times \sqrt{3} [/tex]

[tex]\qquad \sf  \dashrightarrow \:x = 2 \sqrt{3} [/tex]

i hope you you understood

The coordinates of the image after a dilation with respect to the origin is A'' (2,10), B" (0,4), C" (4,2) .

Dilation of a figure will leave its sides get scaled (multiplied) by same number. That number is called the scale factor of that dilation.

Its also called scaling of a figure, but due to the involvement of coordinates, it involves a center of a dilation, which is like a pinned point that stays at the same place after dilation.

Its like we enlarge or shorten the size of the figure (or keep it same, when scale factor = 1).

The vertices of a triangle are A(2,5), B(1, 2), and C(3, 1).

A' = 2(2.5) = (4,10)

B' = 2(1,2) = (2,4)

C' = 2(3,1) = (6,2)

Translation of 2 units left means change in x

A'' = (4-2,10) = (2,10)

B" = (2-2,4) = (0,4)

C" = (6-2,2) = (4,2)

Translation of 1 units up means change in y

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The attached figure represents the image of the triangle after a reflection and a dilation

The vertices are given as:

D(0,2) , E(2,4) , and F(4,0) .

The transformations are given as:

The rule of reflection across the x-axis is

(x,y) -> (x,-y)

So, we have:

D' = (0,-2)

E' = (2,-4)

F' = (4,0)

The rule of dilation by a scale factor of 1.5 is:

(x,y) -> 1.5(x,y)

So, we have:

D'' = 1.5 * (0,-2) = (0,-3)

E'' = 1.5 * (2,-4) = (3,-6)

F'' = 1.5 * (4,0) = (6,0)

See attachment for the image of the triangle

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Answer:

-1 feet

Step-by-step explanation:

below means that your answer will be a negative because sea level is at 0 how many feet below 1 so your answer is -1

Answer:

90 feet

Step-by-step explanation:

40 + 50= 90 feet.

40 feet from the base is 40 feet high and 50 feet up is this equation.

Hope it helps sry if I’m wrong

Answer:

24 ft

Step-by-step explanation:

50 ^ 2 - 40 ^ 2 = b^2

2500 - 1600 = b^2

900 = b^2

√900 = √b^2

24 = b

Answer:

length • width = area

expression:

area • height = (result) • 1/3 = volume

Example:

x (area) • 6 (height) = 6x

6x • 1/3 = 2x(inches)³ or 6x ÷ 3 = 2x(inches)³

Answer:

  (-6)³z²  or  -6³z²

Step-by-step explanation:

An exponent is used to signify the number of times a factor appears in a product.

__

In the expression (-6)·(-6)·(-6)·z·z, the factor (-6) appears 3 times, and the factor z appears 2 times. Written with exponents, this could be ...

  (-6)³z²

_____

Additional comment

We can simplify this by realizing the product of the constant factors will be negative. (There is an odd number of minus signs.) The expression can also be written as ...

  -6³z²

The version with (-6) being cubed is a more direct translation to the use of exponents.

Answer: 36

Step-by-step explanation:

By the triangle midsegment theorem, CE=16, CD=10, and ED=10.

So, the perimeter of triangle CDE is 10+10+16=36

Answer:

The answer is

[tex]2 \cos(100) + i \sin(100) [/tex]

Step-by-step explanation:

This is a complex number,

[tex]a + bi[/tex]

First, convert this to de movire form.

[tex]r( \cos( \alpha ) + i \sin( \alpha ) [/tex]

where

[tex]r = \sqrt{ {a}^{2} + {b}^{2} } [/tex]

and

[tex] \alpha = \tan {}^{ - 1} ( \frac{b}{a} ) [/tex]

[tex]a = 4[/tex]

[tex]b = - 4 \sqrt{ 3} i[/tex]

[tex]r = \sqrt{ {4}^{2} + ( - 4 \sqrt{3}) {}^{2} } [/tex]

[tex]r = \sqrt{16 + 48} [/tex]

[tex]r = \sqrt{64} = 8[/tex]

and

[tex] \alpha = \tan {}^{ - 1} ( \frac{ - 4 \sqrt{3} }{4} ) [/tex]

[tex] \alpha = \tan {}^{ - 1} ( - \sqrt{3} ) [/tex]

Here, our a is positive and b is negative so our angle in degrees must lie in the fourth quadrant, that angle is 300 degrees.

So

[tex] \alpha = 300[/tex]

So our initially form is

[tex]8( \cos(300) + i \sin(300) )[/tex]

Now, we use the roots of unity formula. To do this, we first take the cube root of the modulus, 8,

[tex] \sqrt[3]{8} = 2[/tex]

Next, since cos and sin have a period of 360 we add 360 to each degree then we divide it by 3.

[tex] \sqrt[3]{8} ( \cos( \frac{300 + 360n}{3} ) + \sin( \frac{300 + 360n}{3} ) [/tex]

[tex]2 \cos(100 + 120n) + i \sin(100 + 120n) [/tex]

Since 100 is in the second quadrant, we let n=0,

[tex]2 \cos(100) + i \sin(100) [/tex]