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Luba_88 [7]
2 years ago
6

Does a proportional relationship occur for addition and subtraction or only for multiplication and division?

Mathematics
2 answers:
lord [1]2 years ago
6 0

Answer:

I am pretty sure it is with all of them

Step-by-step explanation:

Studentka2010 [4]2 years ago
5 0
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Family paid $24,000 as a down payment for a home if this represents 12% of the price of the home, find the price of the home.
gizmo_the_mogwai [7]

Answer:

$200,000

Step-by-step explanation:

Down payment = $24,000

Given that $24,000 is 12% of the price.

12% = $24000

1% = 24000 ÷ 12 = $2000

Price of the house

= 2000 x 100%

= $200,000

8 0
3 years ago
Read 2 more answers
Noaya read a book cover to cover in a single session, at a rate of 55 pages per hour. After 4 hours, he had 350 pages left to re
natali 33 [55]

Answer:

55 pages/hour

350 - 55x = y

6 0
3 years ago
Please help.
Charra [1.4K]

Answer:

Since we have a 30° - 60° - 90° triangle, we can calculate any side by knowing at least one out of three:

Since the length of the hypotenuse is twice as long as the shorter leg, we have:

The length of the short leg is: 25/2 = 12.5

Since the length of the longer leg is equal to the length of the shorter leg

multiply by square root of 3 we have:

The length of the longer leg is: 12.5 × √3 ≈21.65

=> The perimeter of the triangle is: 25 + 12.5 + 21.65 = 59.15

6 0
3 years ago
Read 2 more answers
Translate the sentence into an inequality.<br><br> The sum of 7 and w is greater than 21.<br><br> ?
Alexeev081 [22]

Answer:

7+w \ge 21

remove the line under the arrow

7 0
2 years ago
Read 2 more answers
Apply Newton’s method to estimate the solution of x3 − x − 1 = 0 by taking x1 = 1 and finding the least n such that xn and xn +
ss7ja [257]

Solution :

Given

$f(x)=x^3-x-1, x_1=1$

$f'(x)=3x^2-1$

Let the initial approximation is $x_1 =1$

So by Newton's method, we get

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$

$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$

$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$

$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$

$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$

$x_5 \approx x_6$ are identical up to eight decimal places.

The approximate real root is x ≈ 1.32471795

∴ x = 1.32471795

4 0
3 years ago
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