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2 years ago

What is the square root of81

Mathematics

1 answer:

iogann1982 [59]2 years ago

3 0

Answer:

The answer is 9

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Find the area of a triangle with a base 2.6 m and height of 7 m.

nirvana33 [79]

(2.6x7)/2

18.2/2

Area=9.1 m^2

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3 years ago

What is the answer to: 25% of 80?

kicyunya [14]

Answer:

25% of 80

<=> 25/100 of 80

<=> 1/4 of 80

<=> 80/4 = 20

6 0

2 years ago

What is the answer? im trying to figure it out and its hurting my head

Naya [18.7K]

7x-99=2x+1
5x=100
x=20

mark me brainliest if the answer is correct, thank you in advance :)

3 0

3 years ago

Round Ed to the nearthest tenth and answer now question

Vikki [24]

Answer:

$y = 9.1$

Step-by-step explanation:

y can be found using the Law of sines as explained below:

m < Y = 106°

m < X = 58°

WY = x = 8

WX = y = ?

Thus,

$\frac{x}{sin(X)} = \frac{y}{sin(Y)}$

$\frac{8}{sin(58)} = \frac{y}{sin(106)}$

$\frac{8}{0.848} = \frac{y}{0.961}$

Multiply both sides by 0.961 to solve for y

$\frac{8}{0.848}*0.961 = \frac{y}{0.961}*0.961$

$\frac{8*0.961}{0.848} = y$

$\frac{8*0.961}{0.848}*0.961 = y$

$9.07 = y$

$y = 9.1$ (to the nearest tenth)

7 0

3 years ago

Finding Derivatives Implicity In Exercise,Find dy/dx implicity.<br> x2e - x + 2y2 - xy = 0

Klio2033 [76]

Answer:

the question is incomplete, the complete question is

"Finding Derivatives Implicity In Exercise,Find dy/dx implicity . $x^{2}e^{-x}+2y^{2}-xy$ "

Answer : $\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

Step-by-step explanation:

From the expression $x^{2}e^{-x}+2y^{2}-xy$ " y is define as an implicit function of x, hence we differentiate each term of the equation with respect to x.

we arrive at

$\frac{d}{dx}(x^{2}e^{-x )+\frac{d}{dx} (2y^{2})-\frac{d}{dx}xy=0\\$

for the expression $\frac{d}{dx}(x^{2}e^{-x})$ we differentiate using the product rule, also since y^2 is a function of y which itself is a function of x, we have

$(2xe^{-x}-x^{2}e^{-x})+4y\frac{dy}{dx}-x\frac{dy}{dx} -y=0\\\\(2-x)xe^{-x}+(4y-x)\frac{dy}{dx}-y=0 \\$ .

if we make dy/dx subject of formula we arrive at

$(4y-x)\frac{dy}{dx}=y-(2-x)xe^{-x}\\\frac{dy}{dx}=\frac{y-(2-x)xe^{-x}}{(4y-x)}$

5 0

3 years ago