What is 11/14 minus 3/8 =

Mathematics

1 answer:

bixtya [17]4 years ago

6 0

Answer: 23/56

Step by step:11/14 -3/8 =11.4/14.4 - 3.7/8.7 = 44/56

44/56 - 21/56 = 44 - 21/56 = 23/56

You might be interested in

How do you answer this?

gulaghasi [49]

Answer:

75.7°

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relations between trig functions and sides of a right triangle. You are given all three sides of the triangle, so you can make use of at least two different trig functions to find the missing angle.

Cos = Adjacent/Hypotenuse

Tan = Opposite/Adjacent

<h3>cosine</h3>

The hypotenuse is 65, and the side adjacent to the unknown angle is 16. That tells you ...

cos(?) = 16/65

The inverse function is used to find the angle value:

? = arccos(16/65) ≈ 75.7°

<h3>tangent</h3>

The side opposite the angle of interest is 63. Then you have ...

tan(?) = 63/16

The inverse function is used to find the angle value:

? = arctan(63/16) ≈ 75.7°

_____

<em>Additional comments</em>

When using trig functions on a calculator, you need to make sure the angle mode is set to what you want. Here, we want angles in degrees, so we have set that as the angle mode. The [DEG] icon in the lower left corner of the display confirms this.

We can't tell what you're supposed to round the value to. The attachment gives enough digits for you to be able to round to whatever precision you need.

5 0

2 years ago

The Ace Novelty company produces two souvenirs: Type A and Type B. The number of Type A souvenirs, x, and the number of Type B s

Molodets [167]

Answer:

500 type A; 3500 type B

Step-by-step explanation:

The method of Lagrange multipliers can solve this quickly. For objective function f(x, y) and constraint function g(x, y)=0 we can set the partial derivatives of the Lagrangian to zero to find the values of the variables at the extreme of interest.

These functions are ...

$f(x,y)=4x+2y\\g(x,y)=2x^2+y-4$

The Lagrangian is ...

$\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda g(x,y)\\\\\text{and the partial derivatives are ...}\\\\\dfrac{\partial \mathcal{L}}{\partial x}=\dfrac{\partial f}{\partial x}+\lambda\dfrac{\partial g}{\partial x}=4+\lambda (4x)=0\ \implies\ x=\dfrac{-1}{\lambda}\\\\\dfrac{\partial \mathcal{L}}{\partial y}=\dfrac{\partial f}{\partial y}+\lambda\dfrac{\partial g}{\partial y}=2+\lambda (1)=0\ \implies\ \lambda=-2$

$\dfrac{\partial\mathcal{L}}{\partial\lambda}=\dfrac{\partial f}{\partial\lambda}+\lambda\dfrac{\partial g}{\partial\lambda}=0+2x^2+y-4=0\ \implies\ y=4-2x^2\\\\\text{We know $\lambda$, so we can find x and y:}\\\\x=\dfrac{-1}{-2}=0.5\\\\y=4-2\cdot 0.5^2=3.5$