All categories

2 years ago

Two pounds of grapes cost $6.

Mathematics

1 answer:

velikii [3]2 years ago

7 0

Answer:

leftmost empty cell: ⅓

rightmost empty cell: 3

Step-by-step explanation:

✔️If 2 pounds grapes cost $6, then x pounds of grapes cost $1.

Thus;

2 : 6 = x : 1

2/6 = x/1

Multiply both sides by

²/6 × 1 = x/1 × 1

²/6 = x

¹/3 = x

x = ⅓

✔️If 2 pounds grapes cost $6, then 1 pound of grapes cost $x.

Thus;

2 : 6 = 1 : x

²/6 = ¹/x

Cross multiply

2x = 6

Divide both sides by 2

2x/2 = ⁶/2

x = 3

You might be interested in

Find the value of x so that the function has the given value.

nataly862011 [7]

For h(x) = -7x + 10, x =1

For r(x) = 4/5x+ 7 , x = -15

Step-by-step explanation:

in order to find the value of x on which the functions will have given values, we will put the functions equal to the given values

So,

h(x) = -7x + 10; h(x) = 3

$h(x) = -7x+10\\3 = -7x+10\\3-10 = -7x+10-10\\-7 = -7x$

Dividing both sides by -7

$\frac{-7x}{-7} = \frac{-7}{-7}\\x = 1$

r(x) = 4/5x+7; r(x) = -5

$r(x) = \frac{4}{5}x +7\\-5 = \frac{4}{5}x +7\\-5-7 = \frac{4}{5}x\\-12 = \frac{4}{5}x\\x = -12 * \frac{5}{4}\\x = -15$

Keywords: Functions

Learn more about functions at:

#LearnwithBrainly

3 0

3 years ago

Line A is perpendicular to line B. The slope of line A is -1/9 what is the slope of line b?

Pani-rosa [81]

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{-\cfrac{1}{9}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{9}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{9}{1}}\implies 9}$

4 0

3 years ago

Read 2 more answers

I need an answer! NO LINKS PLS! il give brainlist

cricket20 [7]

Answer:

D. The company's chocolate bars weigh 3.2 ounces on average.

Step-by-step explanation:

We are given that a company claims that its chocolate bars weigh 3.2 ounces on average.

The company took many large samples, and each time the mean weight of the sample was within the 95% confidence interval.

Definition of 95% confidence level: 95% confidence level means a range of values that you can be 95% certain contains the true mean of the population.

Thus by considering definition we can conclude that The company's chocolate bars weigh 3.2 ounces on average.

Thus Option D is correct.

D. The company's chocolate bars weigh 3.2 ounces on average.

4 0

2 years ago

The expression (x − 4)2 is equivalent to x2 − 16 x2 + 16 x2 − 8x + 16 x2 + 8x + 16

zzz [600]

Answer:

Given below

Step-by-step explanation:

The algebraic identity is (a-b)^2

= a^2 + b^2 - 2ab

So it'll be,

(x-4)^2

= x^2 + (4)^2 - (2)(x)(4)

= x^2 -8x + 16

or x^2 +16 -8x

4 0

2 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago