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nikitadnepr [17]

3 years ago

6

Carmen bought x candy bars, she game 3 of them away, enter the expression into the box to represent the number of candy bars Car

men has now. PLSSS HELP!!!

1 answer:

anastassius [24]3 years ago

7 0

y=x-3 is the most plausible answer i can give you. Was that the whole question?

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The area of a room is 600 square feet. The length is (x + 5) feet and the width is (x + 4) feet. Find the dimensions of the room

tangare [24]

I'm going to assume that the room is a rectangle.

The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.

You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
$A = lw\\
600 = (x+5)(x+4)\\
600 = x^{2} + 9x + 20
$

Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
$600 = x^{2} + 9x + 20\\
x^{2} + 9x - 580 = 0\\
(x + 29)(x - 20) = 0\\
x + 29 = 0, \:\: x - 20 = 0\\
x = -29, x = 20$

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.

The dimensions of your room are 25ft (length) by 24ft (width).

8 0

4 years ago

Read 2 more answers

How can we find the measure of an exterior angle if we know the measurement of an interior angle

Alik [6]

Step-by-step explanation:

Put the interior angle equal to 180.

It would look something like this:

x+45=180

-45 -45

180-45= 135 so the exterior angle would be 135!

3 0

3 years ago

Point A is at (6,-6) and point C is at(-6,-2). find the coordinates of B on AC so that AB= 3/4AC.

loris [4]

Answer:

B(-6, 0)

Step-by-step explanation:

You want to find B such that ...

(B -A) = (3/4)(C -A) . . . . the required distance relation

4(B -A) = 3(C -A) . . . . . . multiply by 4

4B = 3C +A . . . . . . . . . . add 4A, simplify

Now, we can solve for B and substitute the given coordinates:

B = (3C +A)/4 = (3(-6, -2) +(-6, 6))/4 = (-24, 0)/4 = (-6, 0)

The coordinates of point B are (-6, 0).

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

:D

6 0

3 years ago

Plz answer these questions without putting links and files

kotykmax [81]

Answer:

11/8 and 21/16

Step-by-step explanation:

3 x 1 = 3

4 x 2= 8

3/8+3/4(convert to eighth's)

3/8+6/8=11/8

1 1/8 = 9/8

9 x 1 = 9

8 x 2 = 16

9/16+3/4(convert to sixteenth's)

9/16+12/16

21/16

---

hope it helps

8 0

3 years ago

Read 2 more answers

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