What is the answer to the equation 48-(-31)

Mark went to the sandwich shop. He had a choice of 2 types of bread, 3 meats, and 3 drinks. How many different combinations did

kobusy [5.1K]

B(Bread) D(Drink) M(Meat)

B1+D1+M1
B1+D1+M2
B1+D1+M3
B1+D2+M1
B1+D2+M2
B1+D2+M3
B1+D3+M1
B1+D3+M2
B1+D3+M3

B2+D1+M1
B2+D1+M2
B2+D1+M3
B2+D2+M1
B2+D2+M2
B2+D2+M3
B2+D3+M1
B2+D3+M2
B2+D3+M3

Making there be 18 different combinations.

6 0

3 years ago

Task 1

CaHeK987 [17]

Answer:

part 1 : y = x^2 + 3x + 5

y = x + 13

solution:

x^2 + 3x + 5 = x + 13

x^2 + 2x -8 = 0

(x+4)(x-2) = 0

x = -4 or x = 2

if x = -4, y = 9

if x = 2 , y = 15

part 2 : x^2 + 2x + 1 = 0

(x + 1)^2 = 0

x = -1

y = 1

task 2

part 1 : Let our two digit number be 26

Next, we rewrite it as a difference of two numbers

26=32-6

We want to square 26 using the identity:

In this case,

x=32

y=6

Substituting into the identity

plz make me brainliest

7 0

2 years ago

D=ABC for A solve for A

Dafna1 [17]

A = D/BC. In order to isolate A, all you do is divide by BC

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

2 years ago

Ryan has his own business selling watches, and he wants to monitor his profit per watch sold. The expression 200x-300/x models t

ololo11 [35]

Okay I’ll solve this but it’ll take a while to answer but I’ll try my best to finish as fast as I can!

8 0

2 years ago