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3 years ago

The length of a rectangle is twice the width. the perimeter is 32 cm more than the width. find the dimensions of the rectangle.

1 answer:

7 0

Width: x
Length: 2x
Perimeter=2x+2x+x+x=6x
32 cm more than width means x+32
6x=x+32
5x=32
x=6.4
Dimension is 6.4×12.8 (width×length)

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Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1

lana [24]

Answer:

3 &4

Step-by-step explanation:

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2 years ago

5x+4y = -14<br> 3x+6y=6<br> What is the answer

iren [92.7K]

Hallo! Her er svaret ditt på dette spørsmålet! ☺

Answer: If you are solving by Substitution, your answer will be...

Point form : (-6,4)

Equation form : x=-6 , y=4

Step-by-step explanation: Solve for the first variable in one of the euations, then substitute the result into the other equation.

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3 years ago

Can someone help me please

IgorLugansk [536]

Answer:

Step-by-step explanation:

The slope is 4.

Let's find the slope between your two points.

(1,−2); (3,6)

(x1,y1) = (1,−2)

(x2,y2) = (3,6)

Use the slope formula:

m = y2−y1 / x2−x1

= 6− −2 / 3−1

= 8 / 2

= 4

Answer:

m = 4

7 0

2 years ago

Read 2 more answers

What is the slope of the line that contains the point ( -2, 7) and (2,3)

xz_007 [3.2K]

The slope is 1. The picture will show you why.

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3 years ago

Carlos drew a plan for his garden on a coordinate plane. Rose bushes are located at A(–5, 4), B(3, 4), and C(3, –5)

BartSMP [9]

Given:

A(-5,4)

B(3,4)

C(3,-5)

So point D is:

so point D is (-5,-5)

For AB is

Distance between two point is:

$\begin{gathered} (x_1,y_1)and(x_2,y_2) \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}$

so distance between A(-5,4) and B(3,4) is:

$\begin{gathered} D=\sqrt[]{(3-(-5))^2+(4-4)^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}$

So AB is 8 unit apart.

For B(3,4) and C(3,-5).

$\begin{gathered} D=\sqrt[]{(3-3)^2+(-5-4)^2} \\ =\sqrt[]{0^2+(-9)^2} \\ =9 \end{gathered}$

So BC is 9 unit apart.

For fourth bush point is (-5,-5) it left of point C(3,-5) is:

$\begin{gathered} D=\sqrt[]{(3-(-5))^2+(-5-(-5))^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}$

so fourth bush is 8 unit left of C.

For fourth bush(-5,-5) below to point A(-5,4)

$\begin{gathered} D=\sqrt[]{(-5-(-5))^2+(4-(-5))^2} \\ =\sqrt[]{0^2+9^2} \\ =9 \end{gathered}$

so fourth bush 9 units below of A.

8 0

1 year ago