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3 years ago

7

If a, b, and c are positive integers, what

1 answer:

r-ruslan [8.4K]3 years ago

4 0

The greatest common factor is 2ab

To determine this you need to determine what is divide in both 18 and 8. The only number that can divide in both numbers is 2. Which means they have a common factor/number. And then you continue looking at the expression and see what other numbers or letters are common. The only common thing besides the number is a and b since it’s found in both. There is no c in 18ab.
Hope this makes sense

Please mark as a brainliest (crown) above my comment. Would be much appreciated

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Plzzzz help me on this questions fast <br><br>This is Trigonometry

Sladkaya [172]

Answer:

x ≈ 20.42, y ≈ 11.71

Step-by-step explanation:

Using the cosine ratio on the right triangle on the right, that is

cos20° = $\frac{adjacent}{hypotenuse}$ = $\frac{11}{y}$

Multiply both sides by y

y × cos20° = 11 ( divide both sides by cos20° )

y = $\frac{11}{cos20}$ ≈ 11.71

Using the sine ratio on the right triangle on the left, that is

sin35° = $\frac{opposite}{hypotenuse}$ = $\frac{y}{x}$ = $\frac{11.71}{x}$

Multiply both sides by x

x × sin35° = 11.71 ( divide both sides by sin35° )

x = $\frac{11.71}{sin35}$ ≈ 20.42

5 0

3 years ago

Read 2 more answers

1. Consider the figure below.

Illusion [34]

Answer:

A triangle sums to 180
x-48+2x-164+x= 180
4x-212=180
x= 90

m<A= 16
m<M=90
m<N= 52
Right Triangle
Give brainlist
I do exams and quizzes

7 0

2 years ago

Y=4x+2, what is y-intercept

EleoNora [17]

The y-intercept is (0,2)… Hope this helps..

3 0

2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago

????????????????????????

Tpy6a [65]

Answer:

Show us the question

Step-by-step explanation:

7 0

3 years ago