All categories

2 years ago

Basics test part one

Mathematics

1 answer:

GarryVolchara [31]2 years ago

6 0

Wait so what is the point of this question?

You might be interested in

What is the volume of the prism? m 14 m 6 m 8 m

mestny [16]

Answer:

28?

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

PLEASE HELPPPPPPPPPPPPPPPPPPPPP

astra-53 [7]

Answer:

348

Step-by-step explanation:

Length-l

Width-w

Height-h

A=2(wl+hl+hw)=2·(8·9+6·9+6·8)=348

3 0

3 years ago

the area of a rectangular volleyball court is 1800 square feet the courts length is twice it width write a system of equations t

Marianna [84]

Answer:

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

Step-by-step explanation:

Let

x ----> the length of rectangular volleyball court

y ---> the width of the rectangular volleyball court

we know that

The area of the rectangular volleyball court is equal to

$A=xy$

$A=1,800\ ft^2$

$1,800=xy$ ----> equation A

$x=2y$ -----> equation B

substitute equation B in equation A

$1,800=(2y)y$

$1,800=2y^2$

Solve for y

Simplify

$900=y^2$

take square root both sides

$y=30\ ft$

<em>Find the value of x</em>

$x=2y$

substitute the value of y

$x=2(30)=60\ ft$

therefore

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

7 0

3 years ago

cora arrived at school for 1 hr for doctores appointment she was dismissed from school at 3 15 pm how long was cora at school

Ostrovityanka [42]

Answer:

7 hours

Step-by-step explanation:

5 0

3 years ago

The hypotenuese of an isosceles right triangle is 13 inches. The midpoints of its sides are connected to form an inscribed trian

Georgia [21]

Answer:

The Sum of the areas of theses triangles is 169/3.

Step-by-step explanation:

Consider the provided information.

The hypotenuse of an isosceles right triangle is 13 inches.

Therefore,

$x^2+x^2=169\\2x^2=169\\x=\frac{13}{\sqrt{2} }$

Then the area of isosceles right triangle will be: $A=\frac{1}{2} x^2$

Therefore the area is: $A=\frac{169}{4}$

It is given that sum of the area of these triangles if this process is continued infinitely.

We can find the sum of the area using infinite geometric series formula.

$S=\frac{a}{1-r}$

Substitute $a=\frac{169}{4} \ and\ r=\frac{1}{4}$ in above formula.

$S=\frac{\frac{169}{4}}{1-\frac{1}{4}}$

$S=\frac{\frac{169}{4}}{\frac{3}{4}}$

$S=\frac{169}{3}$

Hence, the Sum of the areas of theses triangles is 169/3.

7 0

3 years ago