state wether the triangles could be proven congruent, if possible, by SSS or SAS. Then, write a congruency statement

McDonalds sells quarter-pound cheeseburgers for .99 cents last saturday.they sold 310 of these cheeseburgers. how many pounds of

Anvisha [2.4K]

The first thing you should do is calculate how many pounds of cheese a hamburger has
1 hamburger = 1/4 pound of cheese.
Then, you can make a rule of three to know how many pounds of cheese were sold:
1 ---> 1/4
310 ---> x
Clearing x we have:
x = ((310) / (1)) * (1/4) = 77.5
Answer
they sell last saturday 77.5 pounds of cheese

4 0

3 years ago

What is equal to n in this equation

IRISSAK [1]

Answer:

n = -45

Step-by-step explanation:

(n-5) /10 = -5

Multiply each side by 10

(n-5) /10 *10= -5*10

n-5 = -50

Add 5 to each side

n-5+5 = -50+5

n = -45

4 0

3 years ago

A strawberry weighs about 1 out of ten an orange. if an orange weighs about 67 grams, about how many grams does a strawberry wei

andre [41]

Answer:

16.75 g

Step-by-step explanation:

Orange's weight = 67 g

Strawberry's weight

= ¼ of the weight of orange

= ¼ × weight of orange

= ¼ × 67 g

= 67 g/4

= 16.75 g

8 0

3 years ago

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

4 years ago

Does anyone know the answer???

GaryK [48]

Answer:the answer would be B

Step-by-step explanation:

the operation is +

6 0

3 years ago

Read 2 more answers