Scientists are studying the temperature on a distant planet. Let y represent the temperature (in degrees Celsius).
Let x represent the height above the surface (in kilometers). Suppose that x and y are related by the equation =y−235x .
Answer the questions below. Note that a change can be an increase or a decrease.

For an increase, use a positive number. For a decrease, use a negative number.

What is the temperature change for each kilometer we go up from the surface?

Answer: c

Step-by-step explanation:

The product of (x-2i)²=x²-4+i4x

The product or multiplication of two complex numbers is also a complex number. The formula for multiplying complex numbers is: (a + ib) (c + id) = (ac - bd) + i(ad + bc).

Given here  (x-2i)² =(x-2i) (x-2i)

                              =x²-4+i4x

Hence, the product is x²-4+i4x

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Answer:

Option a) They are adjacent angles

Step-by-step explanation:

Given is a picture of a graph with angles 1 to 6 around it.

Out of these angle 2 and 5 are right angles.

1 and 2 are adjacent to each other.

They are not complementary because 1+2 not equals 90

They are neither supplementary since sum does not equal 180

They cannot be vertical because they are not formed by intersection of two lines.

Hence only option a is right

Option a) They are adjacent angles

Answer with Step-by-step explanation:

We are given an inequality:

x+2y≤4

We have to determine its correct graph

(since, 0+2×4=8 which is not less than or equal to 4)

Hence, A and C are not the graph of this inequality

Hence, D is also not the graph of this inequality

Hence, correct graph of x+2y≤4 is:

Graph B

Answer:

Step-by-step explanation:

The difference between 14 and 0 is 14, and the difference between 0 and -10 is 10. 14+10=24 for total change. For the average over 8 hours, we have 24/8=3 degrees

Answer:- The area of base is 52900 square meters. The volume is 2654000 cubic meters.

Explanation:-

Base length of right pyramid (square) a =230m

Height of right pyramid = 150m

Area of base of right pyramid[tex]=a^2=(230)^2=52900\ m^2[/tex]

Volume of right pyramid with square base[tex]=\frac{1}{3}a^2h\\=\frac{1}{3}\times52900\times150=2645000\ m^3[/tex]

Thus, the area of base is 52900 square meters and the volume is 2654000 cubic meters.

The area of the base of the pyramid is 52900 m²

The volume of the square base pyramid of Giza is 2645000 m³

The pyramid is a square base pyramid. Therefore,

where

B = base area

h = height

h = 150 m

Therefore,

area of the base = l²

where

l = length of side

area of the base = 230² = 52900 m²

Volume = 1  / 3 × 52900 × 150

volume = 7935000 / 3

volume = 2645000 m³

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The area of smaller triangle is 59 square feet

Two triangles are similar if they have the same ratio of corresponding sides and equal pair of corresponding angles.

The formula for the area of triangle is

[tex]Area = \frac{1}{2} \times base \times \ height[/tex]

According to the given question.

The area of the larger triangle is 105 square feet.

And the one side of the larger triangle and the smaller traingle is 16 and 12 feet respectively.

Suppose the height of thesmaller triangle be x feet and the height of the larger triangle be y feet.

Since, the corresponding edges of similar triangles are proportional.

Therefore,

[tex]\frac{y}{x} = \frac{16}{12}[/tex]

[tex]\implies y = \frac{4}{3} x[/tex]

Also, the area of larger triangle is 105 square feet.

[tex]\implies \frac{1}{2} \times \frac{4}{3} x \times 16 = 105[/tex]

The above euqtaion can be written as

[tex]\implies \frac{1}{2}\times \frac{4}{3} x \times \frac{4}{3} \times 12 = 105[/tex]

[tex]\implies \frac{1}{2} \times (\frac{4}{3} )^{2} \times x \times 12 = 105[/tex]

[tex]\implies \frac{1}{2} \times x \times 12 = 105 \times \frac{9}{16}[/tex]

[tex]\implies \frac{1}{2} \times x \times 12 = 59[/tex]

⇒ Area of smaller triangle  = 59 square feet

Hence, the area of smaller triangle is 59 square feet.

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The length of the third side of the triangle would be 106.5 units approx.

A triangle is a two - dimensional figure with three sides and three angles.

The sum of the angles of the triangle is equal to 180 degrees.

∠A + ∠B + ∠C = 180°

Given is that in triangle ABC, angle B = 30°, a = 210, and b = 164.

The cosine formula is given as follows -

c² = a² + b² + 2abcos(α)

c = √(a² + b² + 2abcos(α))

c = √(210)² + (164)² + 2(210)(164)cos(30°)

c = 106.5 units approx.

Therefore, the length of the third side of the triangle would be 106.5 units approx.

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Answer:

There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]

Step-by-step explanation:

Given : Equation [tex]16x^2=100[/tex]

To find : How many solutions will there be to the following equation?      

Solution :

Equation [tex]16x^2=100[/tex]

Solve the equation,

Divide by 16 both side,

[tex]\frac{16x^2}{16}=\frac{100}{16}[/tex]

[tex]x^2=\frac{100}{16}[/tex]

Taking root both side,

[tex]x=\sqrt{\frac{100}{16}}[/tex]

[tex]x=\sqrt{\frac{10^2}{4^2}}[/tex]

[tex]x=\pm\frac{10}{4}}[/tex]

[tex]x=\pm\frac{5}{2}}[/tex]

Therefore, There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]

Answer:

[tex]x=\frac{7}{3} , \frac{7}{2}[/tex]

Step-by-step explanation:

[tex]6= \frac{35}{x} - \frac{49}{x^2}[/tex]

Now we need to solve for x

To get 'x' alone we make the denominators same

LCD = x^2

WE multiply the whole equation by x^2

[tex]6x^2 = 35x - 49[/tex]

Now we the equation =0, move all the terms to left hand side

[tex]6x^2-35x + 49=0[/tex]

Now we apply quadratic formula to solve for x

a= 6, b= -35 , c= 49

[tex]x= \frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]

[tex]x= \frac{-(-35)+-\sqrt{(-35)^2-4(6)(49)}}{2*6}[/tex]

[tex]x= \frac{35+-\sqrt{49}}{12}[/tex]

[tex]x= \frac{35+-7}{12}[/tex]

Now frame two equations , one with +  and another with -

[tex]x= \frac{35+7}{12}[/tex]                                  [tex]x= \frac{35-7}{12}[/tex]

[tex]x= \frac{42}{12}[/tex]                                       [tex]x= \frac{28}{12}[/tex]  

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

So value of x= {7/3, 7/2}

Answer with explanation:

It is given that, coordinates of Isosceles trapezoid J K L M are J(-b, c), K(b,c), L(a,0), and M(-a,0).

To Prove: The diagonals of an isosceles trapezoid are congruent.

Proof:

  Distance formula , that is distance between two points in x y plane is given by

       [tex]=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]

Where, [tex](x_{1},y_{1}),(x_{2},y_{2})}[/tex] are coordinates of two points in the plane.

Length of Diagonal J L

                  [tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]

Length of Diagonal K M

           [tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]

So, we can see that,

  J L = KM  [tex]=\sqrt{(a+b)^2+(c)^2}[/tex]

Hence,The diagonals of an isosceles trapezoid are congruent.

So ,

  [tex]KM=\sqrt{(a+b)^2+c^2}[/tex]

Option C

The probability that Mr. Calloway will draw a purple piece of paper during the first class period and a blue piece during the second class period with replacement is [tex]\frac{14}{25}[/tex].

The question involves calculating the probability of drawing a purple piece of paper during the first class period and a blue piece of paper during the second class period with replacement. To find this probability, we first calculate the probability of each event separately since the events are independent. Mr. Calloway's hat contains a total of 25 pieces of paper (2 blue, 6 red, 10 yellow, and 7 purple).

The probability of drawing a purple piece of paper in the first class period is the number of purple papers divided by the total number of papers:

P(purple) = Number of purple pieces / Total pieces = [tex]\frac{7}{25}[/tex]

Since the pieces are replaced, the probabilities in the second class period are the same as in the first. Thus, the probability of drawing a blue piece of paper in the second class is:

P(blue) = Number of blue pieces / Total pieces = [tex]\frac{2}{25}[/tex]

Because these events are independent, we multiply the probabilities of each event happening:

Total probability = P(purple)  times P(blue) = ([tex]\frac{7}{25}[/tex])  times ([tex]\frac{2}{25}[/tex]) = [tex]\frac{14}{25}[/tex].

Therefore, the probability that Mr. Calloway draws a purple piece of paper during the first class period and a blue piece of paper during the second class period with replacement is [tex]\frac{14}{25}[/tex].

Answer:

Area will increase by 4 times

Step-by-step explanation:

Given: A rectangle has a base of 3 inches and a height of 9 inches.  

To find: If the dimensions are doubled, what will happen to the area of the rectangle?

Solution:

It is given that a rectangle has a base of 3 inches and a height of 9 inches.

Now, to find if the dimensions are doubled what will happen to the area, first we need to find the original area.

Original area of rectangle, when base is 3 inches and height 9 inches is

[tex]9\times3=27[/tex] square inches

Now, when dimensions are doubled , the base becomes 6 inches and height becomes 18 inches

So, new area becomes [tex]6\times18=108[/tex] square inches.

Now,

[tex]\frac{\text{new area}}{\text{original area} }=\frac{108}{27}[/tex]

[tex]\implies\frac{\text{new area}}{\text{original area} }=\frac{4}{1}[/tex]

Hence, the area will increase by 4 times.

The largest number of goody bags that Kara can make are [tex]18[/tex] .

Highest or greatest Common Factor is the largest common factor that all the numbers have in common.

We have,

Number of Lollipops [tex]=90[/tex]

Number of chocolate bars [tex]=36[/tex]

Number of gumballs [tex]=72[/tex]

So,

To find the number of bags;

First find out the Highest Common Factor of [tex]90,36,72[/tex];

[tex]90=2*3*3*5[/tex]

[tex]36=2*2*3*3[/tex]

[tex]72=2*2*2*3*3[/tex]

So, from the factors of all numbers we have,

Highest Common Factor [tex]=18[/tex]

Now,

Lollipops [tex]=\frac{90}{18} =5[/tex]

Chocolate bars [tex]=\frac{36}{18} =2[/tex]

Gumballs [tex]=\frac{72}{18} =4[/tex]

So, the largest number of goody bags that Kara can make are [tex]18[/tex] so that each goody bag has [tex]5[/tex] number of lollipops, [tex]2[/tex] number of chocolate bars, and [tex]4[/tex] number of gumballs.

Hence, we can say that the largest number of goody bags that Kara can make are [tex]18[/tex] .

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