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3 years ago

Anyone know how to do this

Mathematics

1 answer:

zheka24 [161]3 years ago

4 0

Y-2= -1/2(x+6)
Y-2= -1/2x - 3
Answer is:
Y= -1/2x -1
So your answer is A

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An adult rhino weighed 4,000 pounds. How

amid [387]

B. 2 tons is the ANS

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3 years ago

Read 2 more answers

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.

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3 years ago

Prove each of these identities.

Stella [2.4K]

Answer:

Identity (a) can be re-written as

$sec x\ cosec x - cot x = tan x$

which we already proven in another question, while for idenity (b)

$(A)\frac 1 {sin x} -sin x = cos x \frac{cos x}{sin x}\\\\(B)\frac {1-sin^2x}{sin x} = \frac {cos^2x} {sin x} \\\\(C) \frac {cos^2x}{sin x} =\frac {cos^2x} {sin x}$

step A is simply expressing each function in terms of sine and cosine.

step B is adding the terms on the LHS while multiplying the one on RHS.

step C is replacing the term on the numerator with the equivalent from the pithagorean identity $cos^2x + sin^2x = 1$

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2 years ago

Find the slope of the line perpendicular to the given points. Use / for a fraction. Explain how

stiv31 [10]

Answer:

i think -1/26

Step-by-step explanation:

(y2 - y1) / (x2 - x1)

(-15+16) /(-7-19)

1/-26

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4 years ago

Factor the four-term polynomial by grouping 3x-3+x^3-4x^2

yaroslaw [1]

(x4−3x3+4x2−8)/(x+1) = x3−4x2+8x−8.

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4 years ago