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3 years ago

12 square feet of paper at $1 per square feet

1 answer:

6 0

<span>Given data:
12 square feet of paper at 1 dollars per square feet
=> 12 square feet * 1 dollar = 12 dollars
4 feet of wood at 3 dollars per foot
=> 4 feet of wood * 3 dollars = 12 dollars
14 yards of string for 2 dollars per yard
=> 14 yards of string * 2 dollars = 28 dollars
Now let’s solve that total:
=> 12 dollars + 12 dollars + 28 dollars
=> 52 dollars.
Thus, Maria spent a total of 52 dollars in buying her need supplies.</span>

You might be interested in

A university law school accepts 6 out of every 11 applicants. If the school accepted 564 students, find how many applications th

Elan Coil [88]

I would be 1034 because if you do 6x94 it’s 564 so if you do 11x94 you get the total witch is 1034

5 0

3 years ago

Eugene walked all the way to school at 3 mph then realized she forgot her math book (how could she?!), so she ran back at 7 mph.

Fynjy0 [20]

$\text{Answer: }1\frac{23}{40}\ miles\text{ far she live from school.}$

Explanation:

Since we have given that

Speed at which Eugene walked all the way to school = 3 mph

After realizing that she forgot her math book, she ran back

Speed at which Eugene walked back to her house = 7 mph

Let the distance between her house and her school be x

According to question,

$\frac{x}{3}+\frac{x}{7}=\frac{45}{60}\\\\\frac{7x+3x}{21}=\frac{3}{4}\\\\\frac{10x}{21}=\frac{3}{4}\\\\x=\frac{3\times 63}{4\times 10}=\frac{63}{40}=1\frac{23}{40}\ miles.$

Hence,

$1\frac{23}{40}\ miles\text{ far she live from school.}$

8 0

3 years ago

A 20 cm long thin wire is used to form a square. What is the area of the square?

sergij07 [2.7K]

Answer:

25cm^2

Step-by-step explanation:

A square has 4 equal sides so each side will be 20/4 = 5cm

Easily calculate the area by taking 5 x 5 = 25cm^2

6 0

2 years ago

Solving Rational Functions Hello I'm posting again because I really need help on this any help is appreciated!!

Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

$\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}$

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

$\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

$3 - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x + 3)

$3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$3*(x + 3) - (x + 4) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

$(2*x + 5) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}$

Now we can multiply both sides by (x^2 + 7*x + 12)

$(2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)$

$(2*x + 5)*(x^2 + 7x + 12) = (x + 9)*(x + 4)*(x+3)$

Now we need to solve this:

we will get

$2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)$

$2*x^3 + 19*x^2 + 59*x + 60 = x^3 + 16*x^2 + 42*x + 9$

Then we get:

$2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0$

$x^3 + 3x^2 + 17*x + 51 = 0$

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

The factors of -51 are:

-3 and -17

Let's try -3

p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0

Then x = -3 is one solution of the equation.

But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

$x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0$

The other two solutions are when the other term is equal to zero.

Then the other two solutions are given by:

x = ±√17

And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

6 0

2 years ago

What is the radius of a circle in it in if the area is 36m²?

allsm [11]

Answer:

B. 3.39 m

Step-by-step explanation:

r² = A/π

= 36/3.14

= 11.465

r = √11.465 = 3.39

4 0

2 years ago