Write a number between 80-100 that has 4 factors one of which is 5

. Which ordered pairs in the form (x, y) are solutions to the equation 7x – 5y = 28?

Brut [27]

we know that

If a ordered pair (x,y) is a solution of the equation, then the ordered pair must satisfy the equation

we have the equation

$7x-5y=28$

Let's verify all the cases to determine the solution to the problem.

<u>case A)</u> point $(-6,-14)$

Substitute the values of x and y in the equation

$x=-6\\y=-14$

$7(-6)-5(-14)=28$

$-42+70=28$

$28=28$ -------> is true

therefore

The point $(-6,-14)$ is a solution of the equation

<u>case B)</u> point $(-1,-7)$

Substitute the values of x and y in the equation

$x=-1\\y=-7$

$7(-1)-5(-7)=28$

$-7+35=28$

$28=28$ -------> is true

therefore

The point $(-1,-7)$ is a solution of the equation

<u>case C)</u> point $(4,10)$

Substitute the values of x and y in the equation

$x=4\\y=10$

$7(4)-5(10)=28$

$28-50=28$

$-22=28$ -------> is not true

therefore

The point $(4,10)$ is not a solution of the equation

<u>case D)</u> point $(7,9)$

Substitute the values of x and y in the equation

$x=7\\y=9$

$7(7)-5(9)=28$

$49-45=28$

$4=28$ -------> is not true

therefore

The point $(7,9)$ is not a solution of the equation

therefore

<u>the answer is </u>

$(-6,-14)$

$(-1,-7)$

8 0

3 years ago

Read 2 more answers

If three people are considering registering for a math class, sections: A, B and C:

Artemon [7]

B how many ways can they register for the same section

6 0

2 years ago

There are 59 people going on a field trip.

laila [671]

The answer to this question is the school is going to have to rent 2 vans extra. The reason is because each van hols 8 people right. for the 6 vans that the school already owns, that will be 48 students in vans. When you subract that from 59, you get 11 students that still need a van, so 11 students can not fit in one extra van so the school would need two. Hope this gives you what you were looking for. God bless, and hope this helped. Also, please let me know if it was correct.

6 0

2 years ago

Charlene is a video game designer and wants to make sure that her games can be played on all types of screens. In order for the

natka813 [3]

<h3>Answer:</h3>

C. The playable area has a width of 9 inches and a height of 6 inches.

<h3>Explanation:</h3>

There are a number of ways you can get there.

1. Check the answers to see which have the right area and aspect ratio.

A: area = 24 in² — does not match 54 in²
B: area = 54 in², aspect ratio 6:9 = 2:3 — does not match 3:2 aspect ratio
C: area = 54 in², aspect ratio 9:6 = 3:2 — <em>matches problem statement</em>
D: area = 121.5 in² — does not match 54 in²

2. If the screen were 3:2 (inches), its area would be 6 in². The area of 54 in² is 9 times that value, so the actual screen dimensions are √9 = 3 times 3:2. That is, they are width:height = 9:6 inches — matches selection C.

3. You can write equations for width and height and solve.

w/h = 3/2
wh = 54

Substituting w=3/2·h into the second equation gives

... (3/2)h·h = 54

... h² = 36 . . . . . multiply by 2/3

... h = √36 = 6 . . . . square root, result in inches . . . . matches selection C

3 0

2 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

4 0

3 years ago