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3 years ago

The expression is 3 (x+4)-(2x+7) what is that equivalent too

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

6 0

Answer:

x-19

Step-by-step explanation:

expand the brackets

3x+12-2x-7

3x-2x-12-7

x-19

posledela3 years ago

3 0

Answer:

x+19

Step-by-step explanation:

3x+12 - 2x+7 = x+19

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I need this answered asap !

gtnhenbr [62]

Answer:

Week 16

Step-by-step explanation:

Week 1: 3.5 miles => 22.5 miles left

22.5/1.5 increments = 15 weeks

15 weeks + week 1 = 16th week

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3 years ago

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Which point does the graph of the parent function Y=tan(x) pass through?

zvonat [6]

Answer:

(0, 0)

Step-by-step explanation:

The graph of y = tan x passes through the origin, (0, 0).

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3 years ago

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At a high school the student government is made up of 5 seniors, 4 juniors, 3 sophomores and 2 freshmen. If there are 8 seniors,

hjlf

Answer:

32598720

Step-by-step explanation:

Combination for the school government requires:

5 out of 8 seniors = 8! / 3!5! = 6*7*8/2*3 = 7*8 = 56
4 out of 9 juniors = 9! / 4!5! = 6*7*8*9/2*3*4= 7*2*9 = 126
3 out of 12 sophomores = 12! / 3!9! = 10*11*12/2*3 = 10*11*2 = 220
2 out of 7 fishermen = 7! / 2!5! = 6*7/2 = 21

So number of total ways is:

56*126*220*21 = 32598720

5 0

2 years ago

Please help I don't know what to do ⬇ Find the value of x + 7 when x = 14

ella [17]

Answer:

Step-by-step explanation:

14+7=21

7 0

2 years ago

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Solve the initial value problems: 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1

ladessa [460]

Answer:

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

Step-by-step explanation:

Given the initial value problem $\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\$ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

$\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}$

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}$

The solution to the initial value problem will be;

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

5 0

3 years ago