All categories

3 years ago

Wright 120 percent as a fraction

2 answers:

8 0

$p\%=\dfrac{p}{100}\\\\120\%=\dfrac{120}{100}=\dfrac{12}{10}=\dfrac{12:2}{10:2}=\dfrac{6}{5}=\dfrac{5+1}{5}=\dfrac{5}{5}+\dfrac{1}{5}=1\dfrac{1}{5}\\\\Answer:\ \boxed{120\%=1\dfrac{1}{5}}$

Step2247 [10]3 years ago

7 0

Answer:

1.2 is the decimal, 1 1/5 will be the fraction.

Step-by-step explanation:

Convert the percentage to a fraction by dividing by 100 which leaves us with 1.2

We split the value into two parts, 1 and 0.2

1 is just 1 as a fraction

0.2 is equivalent to 1/5

Together it is 1 1/5

You might be interested in

What is 42 of 100 equivalent to

lilavasa [31]

It is equivalent to 21/50 or 0.42
i got that by dividing two

7 0

3 years ago

Solve for r.<br> P=q+r+s

ddd [48]

In linear equation, The value of the r is a P-q-s.

According to the statement

We have to find that the value of the r .

So, For this purpose, we know that the

The definition of a linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.

From the given information:

The equation is a :

P=q+r+s

This a type of the linear equation then

P=q+r+s

Now, rearrange the terms

P-q-s = r

Then the value of the r is

r = P-q-s .

So, In linear equation, The value of the r is a P-q-s.

Learn more about linear equation here

brainly.com/question/2030026

#SPJ9

4 0

1 year ago

How do i find the area

tatuchka [14]

To find the area its LW or Length times Width.

so you'd do 10x8

10x8=80.

The answer for that question would be 80 yards

8 0

3 years ago

Read 2 more answers

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z +

LUCKY_DIMON [66]

Answer:

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

Step-by-step explanation:

We need to find the directional derivative of the function at the given point in the direction of the vector v.

$f(x, y, z)=xe^{y} + ye^{z} + ze^{x}$ ,point (0, 0, 0) and $v=$

By Theorem: If f is a differentiable function of x , y and z , then f has a directional derivative for any unit vector $\overrightarrow{v} =$ and

$D_{\overrightarrow{u}}f(x,y,z)=f_{x}(x,y,z)u_1+f_{y}(x,y,z)u_2+f_{z}(x,y,z)u_3$

where $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

since, $v=$

then $\overrightarrow{u}=\frac{\overrightarrow{v}}{||v||}$

$\overrightarrow{u}=< \frac{6}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}},\frac{-3}{\sqrt{6^{2}+3^{2}+(-3)^{2}}} >$

$\overrightarrow{u}=< \frac{6}{\sqrt{54}},\frac{3}{\sqrt{54}},\frac{-3}{\sqrt{54}} >$

The partial derivatives are

$f_{x}(x,y,z)=e^{y}+ze^{x}$

$f_{y}(x,y,z)=xe^{y}+e^{z}$

$f_{z}(x,y,z)=ye^{z}+e^{x}$

Then the directional derivative is

$D_{\vec{u}}f(x,y,z)=(e^{y}+ze^{x})(\frac{6}{\sqrt{54}})+(xe^{y}+e^{z})(\frac{3}{\sqrt{54}})+(ye^{z}+e^{x})(\frac{-3}{\sqrt{54}})$

so, directional derivative at point (0,0,0)

$D_{\vec{u}}f(0,0,0)=(e^{0}+0e^{0})(\frac{6}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{3}{\sqrt{54}})+(0e^{0}+e^{0})(\frac{-3}{\sqrt{54}})$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}+\frac{3}{\sqrt{54}}+\frac{-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6+3-3}{\sqrt{54}}$

$D_{\vec{u}}f(0,0,0)=\frac{6}{\sqrt{54}}$

3 0

4 years ago

Number 2 pls, asap! i need this for practice for finals

cestrela7 [59]

Answer:

C. a = 6

Step-by-step explanation:

Use the Pythagorean theorem

$a^{2} + 8^{2} = 10^{2}$

$a^{2}$ + 64 = 100

$a^{2}$ = 36

a = 6

3 0

3 years ago