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3 years ago

9

The perimeter of a standard-sized rectangular rug is 28 ft. the length is 2 ft longer than the width. find the dimensions.

1 answer:

Solnce55 [7]3 years ago

5 0

28=(2+x)+x
-2 -2
_________
26=x+x
26=2x
Divide both by 2
13=x
length=15 width=13

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Divide using Synthetic Division<br> 2. x3 - 4x² + 2x +5<br> X-2

jok3333 [9.3K]

Answer:

the answer is 9

Step-by-step explanation:

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3 years ago

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Pls help me guys!<br> Thanks =P

hichkok12 [17]

I would just try to identify the pattern. First, for the cups column, if the amount increases by 5 each time, then the missing box must be 5 less than 10 or 5

The, for the servings column, you can test to find the pattern. If you were to guess the pattern is 3, when you check, you would see that the boxes are 6 and 9 because then it increases by 3 each time

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Name one landform made by erosion and one landfrom made by deposition. Explain how each landform forms.

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Some landforms made by erosions are:

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3 years ago

Which statement about the values 0.034 and 3.40 are true

Nana76 [90]

Answer:

See Explanation

Step-by-step explanation:

<em>This question requires options.</em>

<em>Since none is provided, I will give a general solution.</em>

<em></em>

We have: 0.034 and 3.40

Required

The relationship between them

3.40 can be expressed as:

$3.40 \to 0.034 * 100$

This means that 3.40 is 100 of 0.034

Similarly, 0.034 can be expressed as:

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3 years ago

Let II be the tangent plane to the graph of f(x, y) = 8 – 2x^2 – 3y^2 at the point (1, 2,-6). Let S, x² + y^2 + z = 4 be another

stealth61 [152]

Let $F(x,y,z)=f(x,y)-z$ . The tangent plane to $f(x,y)$ at (1, 2, -6) has equation

$\nabla F(1,2,-6)\cdot(x-1,y-2,z+6)=0$

We have

$\nabla F(x,y,z)=(-4x,-6y,-1)\implies\nabla F(1,2,-6)=(-4,-12,-1)$

Then the tangent plane has equation

$(-4,-12,-1)\cdot(x-1,y-2,z+6)=0\implies -4(x-1)-12(y-2)-(z+6)=0\implies 4x+12y+z=22$

Let $g(x,y)=4-x^2-y^2$ , and $G(x,y,z)=g(x,y)-z$ . The tangent plane to $S$ at a point $(a,b,c)$ is

$\nabla G(a,b,c)\cdot(x-a,y-b,z-c)=0$

We have

$\nabla G(x,y,z)=(-2x,-2y,-1)\implies \nabla G(a,b,c)=(-2a,-2b,-1)$

so that this plane has equation

$(-2a,-2b,-1)\cdot(x-a,y-b,z-c)=0\implies2ax+2by+z=2a^2+2b^2+c$

In order for this plane to be parallel to the previous plane, we need to have

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so the point we're looking for is (2, 6, -36).

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3 years ago