What is the value of 3 + y ÷ 4, when y = 16?

Higher order thinking: Jeff finds some bugs. He finds 10 Fewer grasshoppers than crickets. He finds 5 fewer crickets than Lady b

Stells [14]

C - 10 = g
L - 5 = C

g = 5

then altogether he finds

40 bugs

8 0

3 years ago

Read 2 more answers

Complete the ordered pair for the given equation. (_,-1);y=-3/5x-7

Fiesta28 [93]

Hi!

So for this problem, we're given the y-coordinate and we can plug that in for y in the equation. So let's do that

-1 = -3/5x - 7
6 = -3/5x
30 = -3x
-10 = x

The missing coordinate would be x, assuming that the equation was y = (-3/5)x - 7

3 0

3 years ago

35 POINTS. I need help! Please!

topjm [15]

There is no solution for the first one

if you eliminate y you get 2 equations
-13x - 13z = -25
-13x - 13x = -15

- there is no solution to theses

a23 means the element in the second row and the 3rd column

so its -5

7 0

3 years ago

Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units. What would be the perimeters of the

Dmitriy789 [7]

Answer:

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

Step-by-step explanation:

Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.

To find - What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?

Proof -

We know that

Are of Rectangle = Length × Breadth

And Perimeter of Rectangle = 2 (Length + Breadth )

Now,

Given that,

Orbet used 24 square tiles

And perimeter of the rectangle made = 20

So,

Possible Length of rectangle = 6

Possible breadth of Rectangle = 4

Or Vice-versa.

Total number of Rectangles possible = 4

Possibilities are - 1 × 24, 2 × 12, 3 × 8, 4 × 6

Case I :

Length of rectangle = 1

Breadth of rectangle = 24

∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units

Area of rectangle = 1 × 24 = 24 unit²

Case II :

Length of rectangle = 2

Breadth of rectangle = 12

∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units

Area of rectangle = 2 × 12 = 24 unit²

Case III :

Length of rectangle = 3

Breadth of rectangle = 8

∴ Perimeter of rectangle = 2(3 + 8) = 2(11) = 22 units

Area of rectangle = 3 × 8 = 24 unit²

Case IV :

Length of rectangle = 4

Breadth of rectangle = 6

∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units

Area of rectangle = 4 × 6 = 24 unit²

∴ we get

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

5 0

3 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

3 years ago