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3 years ago

5

B: identify the constant(s), variable(s), base, exponent, literal and numeral coefficient in each expression.complete the table

1 answer:

docker41 [41]3 years ago

8 0

Answer:

2. 3+2b

Step-by-step explanation:

mark me brainliest pl

hope it help you

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Find the area of a plane figure bounded by lines

natulia [17]

Answer: 4.5

<u>Step-by-step explanation:</u>

First, find the points of intersection by solving the system.

y = x² + 2x + 4

y = x + 6

Solve by substitution:

x² + 2x + 4 = x + 6 ⇒ x² + x - 2 = 0 ⇒ (x + 2)(x - 1) = 0 ⇒ x = -2, x = 1

Now, integrate from x = -2 to x = 1

$\int\limits^1_2 {(x+6)-(x^{2}+2x+4) } \,$ <em>the bottom of the integral is -2 </em>

= $\int\limits^1_2 {x+6-x^{2}-2x-4 } \,$

= $\int\limits^1_2 {-x^{2}-x+2 } \,$

= $\frac{-x^{3}}{3} - \frac{x^{2}}{2}+2x\int\limits^1_2 {} \,$

= $(\frac{-1^{3}}{3} - \frac{1^{2}}{2}+2(1)) - (\frac{-(-2)^{3}}{3} - \frac{(-2)^{2}}{2}+2(-2))$

= $(\frac{-1}{3} - \frac{1}{2} +2) - (\frac{8}{3} -\frac{4}{2} -4)$

= $\frac{-9}{3} + \frac{3}{2} +6$

= -3 + 1.5 + 6

= 4.5

3 0

3 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

4 years ago

Which equations best shows 48 is a multiple of 3

german

3 = 48 divided by 16

Mark brainliest please

Hope this helps you

6 0

3 years ago

Read 2 more answers

Anybody know the answer to this?

avanturin [10]

Answer:

The answer is 18

Step-by-step explanation:

Substitute b for -3 and y for 5 so 3*5=15

You then subtract -3 from 15 so a negative minus a negative is a positive so the equation would be 3*5=15--3= 3*5=15+3=18

8 0

3 years ago

Read 2 more answers

You and 2 friends share 7 cookies equally how many cookies do you each get

UkoKoshka [18]

Well 7/2=3.5 so i would say 3/1/2 i believe

8 0

4 years ago