4) Which expression is equivalent to 3x + 5x? X + 8 8x + 2. 8x. 2x

Mathematics

1 answer:

____ [38]1 year ago

6 0

Answer:

Step-by-step explanation:

3x + 8x are like terms which means they have the exact same variable.

To add like terms we add the coefficients which are 3 and 5.

3 + 5 = 8

We have 8 x's. So 3x + 5x = 8x.

In order to add terms, the variable part has to be exactly the same.

It's like saying 3 balls + 5 balls = 8 balls.

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Given: m

marissa [1.9K]

Answer:

m∠LTE = 110

Step-by-step explanation:

1. add up all of the arcs.

2x+3x+4x-8+x-12

2. all of the arcs equal 360

2x+3x+4x-8+x-12=360

3. Find x

10x-20=360, x=38

4. angle LTE is equal to half of the sum of the intercepted arcs.

0.5(arc LE +GF)

5. plug in LE +GF with x

.5(76+144)

7 0

2 years ago

Read 2 more answers

On Sunday, a local hamburger shop sold a combined of 572 hamburger and cheeseburger. The number of cheeseburgers sold was three

igor_vitrenko [27]

Answer: 143 hamburgers and 429 cheese burgers

Explanation:

Call h and c the number of both items.
(h-hamburger and c-cheeseburger)

h + c = 572
c = 3h

Sub the second into the first

h + 3h = 572
4h = 572

Divide both sides by 4
h = 143 hamburgers

Use this back into the second equation
c = 3 • 143 = 429 cheeseburgers

3 0

2 years ago

Solve the initial value problems:<br> 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;

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

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