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1 year ago

Someone help me thanks! it's on pythagoras theorem btw

Mathematics

1 answer:

Amanda [17]1 year ago

6 0

Answer:

y = 3

Step-by-step explanation:

Pythag theorem : ( only applies to RIGHT triangles)

Hypotenuse ^2 = leg1 ^2 + leg 2^2

(7y-1)^2 = (5y-3)^2 + (5y+1)^2

49 y^2 - 14y +1 = 25y^2 -30 y + 9 + 25y^2 + 10y + 1

-y^2 + 6y -9 = 0

or y^2 -6y + 9 = 0 solve by factoring or using Quadratic Formula

( y -3)(y-3) = 0 shows y = 3

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What’s the volume of one cylinder?

r-ruslan [8.4K]

Answer:

For the first one its: Volume = 402.12in³

For the second one its: Volume = 201.0619 in³

Step-by-step explanation:

For the first one work:

Volume = 3.1416 x 42 x 8

= 3.1416 x 16 x 8

For the second one work:

Volume = 3.1416 x 42 x 4

= 3.1416 x 16 x 4

5 0

3 years ago

Read 2 more answers

1. Given that 4x²- 16x + 15 = a(x-p)²+ q for all values of x.

FinnZ [79.3K]

Answer:

a = 4, p = 2, q = - 1

Step-by-step explanation:

Expand the right side of the identity, then compare the coefficients of like terms with those on the left side.

a(x - p)² + q ← expand (x - p)² using FOIL

= a(x² - 2px + p²) + q ← distribute parenthesis

= ax² - 2apx + ap² + q

Compare coefficients of x² term

a = 4

Compare coefficients of x- term

- 2ap = - 16, that is

- 2(4)p = - 16

- 8p = - 16 ( divide both sides by - 8 )

p = 2

Compare constant terms

ap² + q = 15 , that is

4(2)² + q = 15

16 + q = 15 ( subtract 16 from both sides )

q = - 1

Thus a = 4, p = 2, q = - 1

8 0

3 years ago

Simplify the expression

liberstina [14]

$\frac{5}{x^-2y^5}$
$= \frac{5x^2 }{y^5}$ (Answer C)

8 0

3 years ago

Two hoses are filling a pool the first hose fills at a rate of x gallons per minute the second hose fills at a rate of 15 gallon

Zielflug [23.3K]

Answer:

B. (0, 5]∪(15,30] only (15,30] contains viable rates for the hoses.

Step-by-step explanation:

The question is incomplete. Find the complete question in the comment section.

For us to meet the pool maintenance company's schedule, the pool needs to fill at a combined

rate of at least 10 gallons per minute. If the inequality represents the combined rates of the hoses is 1/x+1/x-15≥10 we are to find all solutions to the inequality and identifies which interval(s) contain viable filling rates for the hoses. On simplifying the equation;

$\frac{1}{x} + \frac{1}{x-15} \geq \frac{1}{10}\\\\ find\ the \ LCM \ of \ the function \ on \ the \ LHS\\\\\frac{x-15+x}{x(x-15)} \geq \frac{1}{10}\\\\\frac{2x-15}{x(x-15)} \geq \frac{1}{10}\\\\10(2x-15)\geq x(x-15)\\\\20x-150\geq x^2-15x\\\\collect \ like \ terms\\-x^2+20x+15x - 150\geq 0\\$

$-x^2+35x-150 \geq 0\\\\multipply \ through \ by \ minus\\x^2-35x+150 \leq 0\\\\(x^2-5x)-(30x+150) \leq 0\\\\x(x-5)-30(x-5) \leq 0\\\\$

$(x-5)(x-30) \leq 0\\\\x-5 \leq 0 and x - 30 \leq 0\\\\x \leq 5 \ and \ x \leq 30$

The interval contains all viable rate are values of x that are less than 30. The range of interval is (0, 5]∪(15,30]. Since the pool needs to fill at a combined rate of at least 10 gallons per minute for the pool to meet the company's schedule, this means that the range of value of gallon must be more than 10, hence (15, 30] is the interval that contains the viable rates for the hoses.

6 0

3 years ago

What is the input value other than 7 for which g(x)=4

Vsevolod [243]

Given:

It is given that the value of the graph when the input 7 is $g(x)=4$

We need to determine the value of x when $g(x)=4$

Value of x when $g(x)=4$ :

The value of x can be determined by using the graph.

From the graph, we need to determine the value of x when $g(x)=4$ other than the value x = 7.

This can be determined by finding the point at which the line meets the point y = 4, we can find the corresponding x - value.

Thus, from the graph, it is obvious that the graph also meets the point y = 4 when x = -8.

Therefore, the input value is x = -8 which makes $g(x)=4$

Hence, the input value other than 7 for which $g(x)=4$ is x = -8.

7 0

3 years ago

Someone help me thanks! it's on pythagoras theorem btw ​

Someone help me thanks! it's on pythagoras theorem btw