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3 years ago

12

PLEASE HELP IM BEING TIMED !! Find the perimeter of a quadrilateral with the given vertices: M(-1, 7), A(5, 2), T(-3, -3), H(0,

5)

1 answer:

allsm [11]3 years ago

5 0

Answer:

28.02

Step-by-step explanation:

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Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

masha68 [24]

Answer:

The Width of the garden will be 9 feet.

Step-by-step explanation:

Given:

Manny has wood =48 feet

Length of the garden = 15 feet

The equation which can be used .

We need to find the width of the garden.

Solution :

Now Length of the garden is given we will substitute in the given equation to find the width.

Now Using Subtraction property we will subtract both side by 30 we get;

Now Using Division Property we will divide both side by 2 we get;

Hence the Width of the garden will be 9 feet.

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3 years ago

Find the volume of the pyramid. in.3

Papessa [141]

Answer:

1.5

Step-by-step explanation:

5 0

3 years ago

If 3/4 = x/8 then x is equal to what

pickupchik [31]

Answer:

x is 6

Step-by-step explanation:

3/4 * x/8

3*8=24 then,

divide by 4 which gives you 6.

Also if you simplify 6/8, it will give you 3/4.

3 0

3 years ago

Read 2 more answers

Solve the following equations;-<br><br> x - 9 /√x + 3 =1

fgiga [73]

Answer:

13 , 6.

Step-by-step explanation:

A equation is given to us and we need to find out the value of x . The given equation to us is ,

$\sf\implies \dfrac{ x -9}{\sqrt{x+3}}= 1$

Cross multiply ,

$\sf\implies x - 9 =\sqrt{ x +3}$

Squaring both sides ,

$\sf\implies (x - 9)^2 = x + 3$

Simplify the whole square ,

$\sf\implies x^2+9^2-2.9.x = x +3 \\$

$\sf\implies x^2+81 - 18x = x -3$

Add 3 and subtract x on both sides ,

$\sf\implies x^2 -18x-x +81-3=0$

Simplify ,

$\sf\implies x^2 -19x +78=0$

Split the middle term of the quadratic equation,

$\sf\implies x^2-13x-6x+78=0$

Take out common ,

$\sf\implies x( x -13)-6(x-13)=0$

Take out ( x -13 ) as common ,

$\sf\implies ( x -13)(x-6)=0$

Equate both factors to 0 ,

$\sf\implies \boxed{\pink{\sf x = 13,6}}$

<u>Hence</u><u> the</u><u> </u><u>value</u><u> of</u><u> </u><u>x</u><u> is</u><u> </u><u>1</u><u>3</u><u> </u><u>or </u><u>6</u><u> </u><u>.</u>

7 0

3 years ago

Is the equation y = x3 even, odd, or neither?

ohaa [14]

Let be y =f(x)= x3, f(-x) = (-x)^3 = - x^3 = - f(-x), f(-x) = (-x)^3, f(-x) = - f(<span>-x)
the equation is odd</span>

5 0

3 years ago