PLEASE HELP PICTURE SHOWN!!

Answers

Answer 1

Answer: The correct answer is d

Answer 2

Answer: = 6^(1/4+1/4+1/4+1/4)
=6^(4/4)
=6^1
= 6
answer is D. 6

Related Questions

The slope of the line whose equation is 5 x - 3 y = 4 is 5/3 4/5 -3/5
PLEASE HELP THIS IS TIMED !!!!!Use the graph of f(x) to explain the relationship between the real zeros of f(x) and its intercept(s).A. ) f(x) has one real zero at –2 because the graph of the function has an intercept at (0, –2).B. ) f(x) has two real zeros at –4 and –2 because the graph of the function has intercepts at (–4, 0) and (0, –2).C. ) f(x) has no real zeros because the graph of the function does not pass through (0, 0).D. ) f(x) has one real zero at –4 because the graph of the function has an intercept at (–4, 0).
What is $40 with a 30% markup?
The owner of a discount clothing store ordered six belts and eight hats for $140. Aweek later, at the same prices, he bought nine belts and six hats for $132. Find theprice of a belt and a hat.
What does 111,111,111×111,111,111 equal

A line passes through the points (-1, 10) and (3, 2). Which shows the graph of this line?8
8
4
2
- 108
2
46
8 10
X
12
-8
-10
10
6

Answers

Answer:

Step-by-step explanation:

EDGE 2021

The cost of 3 kg of apples and 6 kg of oranges is Rs 300. If the cost of 1 kg of oranges is Rs 30, find the cost of 8 kg of apples.

Answers

Answer:

Cost of 1 = 30

So cost of 8kg apples=8*30

=240

Step-by-step explanation:

Is the number rational or irrational?Column A Column B
1. 3/7 A. Rational
2. 0.78( repeating) B. Irrational
3. Square root 7/9
4. 5 pi

Answers

1 and 2 are both rational.
3 and 4 are both not.

lily is making granola she uses almonds walnuts and pecans in tge ratio 4:3:8 lily uses 40ounces of pecans

Answers

Well. im not 100% sure if this is correct. but what i did was did 5 x 8. which equals 40. so then i did 5 x 3. which equals 15. then i did 5 x 4. which equals 20. hopefully im correct. hope this helped! :)

Gina, Sam, and Robby all rented movies from the same video store. They each rented some dramas, comedies, and documentaries. Gina rented 11 movies total. Sam rented twice as many dramas, three times as many comedies, and twice as many documentaries Gina. He rented 27 movies total. If Robby rented 19 movies total with the same number of dramas, twice as many comedies, and twice as many documentaries as Gina, how many movies of each type did Gina rent? Answers:
3 dramas, 5 comedies, and 3 documentaries
2 dramas, 6 comedies, and 3 documentaries
1 dramas, 4 comedies, and 6 documentaries
4 dramas, 3 comedies, and 4 documentaries

Answers

Answer:

Gina rent 3 dramas, 5 comedies, and 3 documentaries.

Step-by-step explanation:

Let Gina rented x movies of dramas , y movies of comedies and z movies of documentaries then , Gina rented total 11 movies.

$\Rightarrow x+y+z=11$ .........(1)

Also given Sam rented twice as many dramas, three times as many comedies, and twice as many documentaries Gina, thus he rented 2x movies of dramas , 3y movies of comedies and 2z movies of documentaries. also, he rented a total of 27 movies.

$\Rightarrow 2x+3y+2z=27$ ............(2)

Also, Robby rented the same number of dramas, twice as many comedies, and twice as many documentaries as Gina, thus, he rented x movies of dramas , 2y movies of comedies and 2z movies of documentaries also, he rented a total of 19 movies.

$\Rightarrow x+2y+2z=19$ ............(3)

Solving the three equation using matrix form,

$\left[\begin{array}{ccc}1&1&1\n2&3&2\n1&2&2\end{array}\right] \left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c}11\n27\n19\end{array}\right]$

This, system is in form of AX= b,

Where, $A=\left[\begin{array}{ccc}1&1&1\n2&3&2\n1&2&2\end{array}\right]$ , $X=\left[\begin{array}{c}x\ny\nz\end{array}\right]$ , $b=\left[\begin{array}{c}11\n27\n19\end{array}\right]$

Pre-mutiply by A inverse both sides,

$X=A^(-1)b$ ............(P)

First finding inverse,

$\mathrm{Augment\:with\:a}\:3x3\:\mathrm{identity\:matrix}$

$=\begin{bmatrix}1&1&1&\mid \:&1&0&0\n 2&3&2&\mid \:&0&1&0\n 1&2&2&\mid \:&0&0&1\end{bmatrix}$

$\mathrm{Swap\:matrix\:rows:}\:R_1\:\leftrightarrow \:R_2$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 1&1&1&\mid \:&1&0&0\n 1&2&2&\mid \:&0&0&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-(1)/(2)\cdot \:R_1$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 1&2&2&\mid \:&0&0&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3-(1)/(2)\cdot \:R_1$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 0&(1)/(2)&1&\mid \:&0&-(1)/(2)&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3+1\cdot \:R_2$

$=\begin{bmatrix}2&3&2&\mid \:&0&1&0\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-2\cdot \:R_3$

$=\begin{bmatrix}2&3&0&\mid \:&-2&3&-2\n 0&-(1)/(2)&0&\mid \:&1&-(1)/(2)&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_2\:\leftarrow \:-2\cdot \:R_2$

$=\begin{bmatrix}2&3&0&\mid \:&-2&3&-2\n 0&1&0&\mid \:&-2&1&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-3\cdot \:R_2$

$=\begin{bmatrix}2&0&0&\mid \:&4&0&-2\n 0&1&0&\mid \:&-2&1&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

$\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_1\:\leftarrow (1)/(2)\cdot \:R_1$

$=\begin{bmatrix}1&0&0&\mid \:&2&0&-1\n 0&1&0&\mid \:&-2&1&0\n 0&0&1&\mid \:&1&-1&1\end{bmatrix}$

Thus, $A^(-1)=\begin{pmatrix}2&0&-1\n -2&1&0\n 1&-1&1\end{pmatrix}$

Put values in equation (P),

$X=A^(-1)b$

$\left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c,c,c}2&0&-1\n -2&1&0\n 1&-1&1\end{array}\right]\left[\begin{array}{c}11\n27\n19\end{array}\right]$

$\left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c,c,c}2\cdot \:11+0\cdot \:27+\left(-1\right)\cdot \:19\n \left(-2\right)\cdot \:11+1\cdot \:27+0\cdot \:19\n 1\cdot \:11+\left(-1\right)\cdot \:27+1\cdot \:19\end{array}\right]$

$\left[\begin{array}{c}x\ny\nz\end{array}\right]=\left[\begin{array}{c}3\n5\n3\end{array}\right]$

Thus, Gina rent 3 dramas, 5 comedies, and 3 documentaries.

3 dramas, 5 comedies, and 3 documentaries

The distances (y), in miles, of two cars from their starting points at certain times (x), in hours, are shown by the equations below:Car A
y = 60x + 10

Car B
y = 40x + 70

After how many hours will the two cars be at the same distance from their starting point and what will that distance be?

2 hours, 150 miles
2 hours, 190 miles
3 hours, 150 miles
3 hours, 190 miles

Answers

when the cars are at the same distance from the starting point the y will be the same for both of the equations.
which means that 60x + 10 = 40x + 70

⇒ 60x-40x=70-10
⇒ 20x=60
⇒ x=3 hours

now that we know the x we can choose an equation and solve it

60x3+10=y⇒
⇒y=180+10
⇒y=190 miles

the correct answer is the last one: 3 hours, 190 miles

Other Questions

Which of the following are polynomials? A- a^0 B- 0 C- a^0 + 17

A kite is a quadrilateral with two pairs of adjacent, congruent sides. The vertex angles are those angles in between the pairs of congruent sides. Prove the diagonal connecting these vertex angles is perpendicular to the diagonal connecting the non-vertex angles. Be sure to create and name the appropriate geometric figures. This figure does not need to be submitted

Hiro is creating a smaller scaled replica of a triangular canvas. Which of the following expressions will help him determine the length of segment AB? AB = AD AB = ACAB= AC times AE/ADAB= AE times AD/AB

Why are cubes and spheres called solid figures?