All categories

3 years ago

What is the value of 1 + x ÷ 5, when x = 80?

Mathematics

2 answers:

Tom [10]3 years ago

4 0

I hope this helps you

1+80÷5

1+16

17

GenaCL600 [577]3 years ago

3 0

17. 80 divided by 5 equals 16. So that +1 would make it 17.

You might be interested in

Select all the correct equations.

anyanavicka [17]

Answer:

that's the answer

Step-by-step explanation:

#platolivesmatter

6 0

2 years ago

Halla la tasa de variación de cada funcion en el intervalo [-4,3] e indica si es positiva , negativa o nula A) f(x)=x2-2x+4 B) f

masya89 [10]

Answer:

$A) \hspace{3}Rate\hspace{3}of\hspace{3}change=-5\hspace{3}Negative\\\\B)\hspace{3}Rate\hspace{3}of\hspace{3}change=-21\hspace{3}Negative$

Step-by-step explanation:

Given a function $f(x)$ , we called the rate of change to the number that represents the increase or decrease that the function experiences when increasing the independent variable from one value " $x_1$ " to another " $x_2$ ".

The rate of change of $f(x)$ between $x_1$ and $x_2$ can be calculated as follows:

$Rate\hspace{3}of\hspace{3}change=f(x_2)-f(x_1)$

For:

$f(x)=x^2-2x+4$

Let's find $f(x_1)$ and $f(x_2)$ , where:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=(-4)^2-2(4)+4=16-8+4=12\\f(x_2)=f(3)=(3)^2-2(3)+4=9-6+4=7$

So:

$Rate\hspace{3}of\hspace{3}change =7-12=-5\hspace{3}Negative$

And for:

$f(x)-3x+2$

Let's find $f(x_1)$ and $f(x_2)$ , where:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

So:

$Rate\hspace{3}of\hspace{3}change =-7-14=-21\hspace{3}Negative$

<em>Translation:</em>

Dada una función $f(x)$ , llamábamos tasa de variación al número que representa el aumento o disminución que experimenta la función al aumentar la variable independiente de un valor " $x_1$ " a otro " $x_2$ ".

La tasa de variación de $f(x)$ entre $x_1$ y $x_2$ , puede ser calculada de la siguiente forma:

$Tasa\hspace{3}de\hspace{3}variacion=f(x_2)-f(x_1)$

Para:

$f(x)=x^2-2x+4$

Encontremos $f(x_1)$ y $f(x_2)$ , donde:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

Entonces:

$Tasa\hspace{3}de\hspace{3}variacion =7-12=-5\hspace{3}Negativa$

Y para:

$f(x)-3x+2$

Encontremos $f(x_1)$ y $f(x_2)$ , donde:

$[x_1,x_2]=[-4,3]$

$f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7$

Entonces:

$Tasa\hspace{3}de\hspace{3}variacion=-7-14=-21\hspace{3}Negativa$

8 0

3 years ago

You and your friend mix water and citric acid. You add 3 cups of citric acid for every 16 cups of water. Your friend adds 2 cups

Viktor [21]

Answer:

You:If it’s was 6 cups of citric I’d be to 32 water

Your friend: if it was 4 cups it’d be 12 water

Step-by-step explanation:

I believe your friends is more acidic because your Is being diluted more

4 0

2 years ago

In the figure shown to the right, two angle measurements are given. Determine the others.

creativ13 [48]

The figure is shown below

From the figure

Angle 150 degree and angle p forms angles on a straight

Since the sum of angles on a straight line equals 180 degrees

Hence

$150^{\circ}+p=180^{\circ}$

Solve for p in the equation

$\begin{gathered} p=180^{\circ}-150^{\circ} \\ p=30^{\circ} \end{gathered}$

Hence, p = 30

From the figure

Angle p and angle q are vertically opposite angles

Since vertically opposite angles are equal then

$p=q=30^{\circ}$

Hence, q = 30

Applying the rule of angles on a straight line

This implies

$q+w+60=180$

Substitute q = 30 into the equation

$30+w+60=180$

Solve for w

$\begin{gathered} w+90=180 \\ w=180-90 \\ w=90 \end{gathered}$

Hence, w = 90

8 0

1 year ago

plain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = x2 − 2x x2 − 4 if x ≠ 2 1 if x =

ElenaW [278]

Answer with Step-by-step explanation:

We are given that

$f(x)=\left\{\begin{matrix}\dfrac{x^2-2x}{x^2-4}&,if\ \ x\neq 2 \\ 1&,if\ \ x=2\end{matrix}\right.$

We have to explain that why the function is discontinuous at x=2

We know that if function is continuous at x=a then LHL=RHL=f(a).

$f(x)=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}$

LHL=Left hand limit when x <2

Substitute x=2-h

where h is small positive value >0

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2-h}{2-h+2}$

$\lim_{h\rightarrow 0}\frac{2-h}{4-h}=\frac{2}{4}=\frac{1}{2}$

Right hand limit =RHL when x> 2

Substitute

x=2+h

$\lim_{h\rightarrow 0}f(x)=\lim_{h\rightarrow 0}\frac{2+h}{2+h+2}=\lim_{h\rightarrow 0}\frac{2+h}{4+h}$

$=\frac{2}{4}=\frac{1}{2}$

LHL=RHL= $\frac{1}{2}$

f(2)=1

$LHL=RHL\neq f(2)$

Hence, function is discontinuous at x=2

4 0

3 years ago