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3 years ago

The number of laughs (denoted by

Mathematics

1 answer:

Sonja [21]3 years ago

6 0

Answer:

jokes . knowledge2

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familiarity

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What is equivalent to 24/16

gulaghasi [49]

We can convert 24/16 to a mixed number. It will be equivalent to 24/16.

24/16
24 ÷ 16 = 1.5
16 can go in to 24 once.
24 - 16 = 8

1 = whole number
8 = numerator
16 = denominator

Lets write our mixed number:-

$1 \frac{8}{16}$

This can be simplified ^^^^

$\frac{8/8}{16/8} = \frac{1}{2}$

So, our old mixed number simplified is $1 \frac{1}{2}$

So, 24/16 is equivalent to 1. $1 \frac{8}{16}$ , 2. $1 \frac{1}{2}$

Hope I helped ya!!! xD

6 0

3 years ago

What is the value of 4 to the power of 3 divided by 2 to the power of 3

sp2606 [1]

Answer:

Step-by-step explanation:

4^3/2^3

(4^3)/(2^3)

4^3 = 64
2^3 = 8

64/8 = 8

7 0

3 years ago

Read 2 more answers

Find the minimum and maximum of f(x, y, z) = y + 4z subject to two constraints, 2x + z = 4 and x2 + y2 = 1. g

yuradex [85]

$L(x,y,z,\lambda_1,\lambda_2)=y+4z+\lambda_1(2x+z-4)+\lambda_2(x^2+y^2-1)$

$L_x=2\lambda_1+2\lambda_2 x=0\implies\lambda_1+\lambda_2x=0$
$L_y=1+2\lambda_2y=0$
$L_z=4+\lambda_1=0\implies\lambda_1=-4$
$L_{\lambda_1}=2x+z-4=0$
$L_{\lambda_2}=x^2+y^2-1=0$

$\lambda_1=-4\implies \lambda_2x=4\implies\lambda_2=\dfrac4x$
$1+2\lambda_2y=0\implies\lambda_2y=-\dfrac12\implies8y=-x$

$x^2+y^2=1\implies (-8y)^2+y^2=65y^2=1\implies y=\pm\dfrac1{\sqrt{65}}$
$y=\pm\dfrac1{\sqrt{65}}\implies x=\mp\dfrac8{\sqrt{65}}$
$2x+z=4\implies z=4\pm\dfrac{16}{\sqrt{65}}$

We have two critical points to consider: $\left(-\dfrac8{\sqrt{65}},\dfrac1{\sqrt{65}},4+\dfrac{16}{\sqrt{65}}\right)$ and $\left(\dfrac8{\sqrt{65}},-\dfrac1{\sqrt{65}},4-\dfrac{16}{\sqrt{65}}\right)$ .

At these points, we respectively have a maximum of $16+\sqrt{65}$ and a minimum of $16-\sqrt{65}$ .

6 0

3 years ago

Please help me with this

saul85 [17]

Answer:

B,C,E (last one)

Step-by-step explanation:

3 0

3 years ago

You can approximate e by substituting large values for ninto the expression

ratelena [41]

You can approximate e by substituting large values for (1 + 1/n)^n into the expression.

<h3>How to illustrate the expression?</h3>

It should be noted that an expression can simply be used to illustrate the information given in a data.

In this case, you can approximate e by substituting large values for (1 + 1/n)^n into the expression.

Learn more about expressions on:

brainly.com/question/723406

#SPJ1

8 0

2 years ago