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3 years ago

NEED HELP ASAP!!! The table shows the relationship between the number of bracelets made and the number of beads used:

1 answer:

5 0

Answer: 49
reason:
okay so basically to get to 3 to 5 you have to add 2. To get to 21 to 35 you need to add 14. To get to 5 to 7 you need to add 2, therefore adding 14 to 35. It's a pattern :)

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If you need 8 oz of chocolate chips to make 1.6 lb of cookie dough how many ounces of chocolate chips will you need to make 7 lb

iren [92.7K]

Answer:

3.5 Oz of chocolate

Step-by-step explanation:

1.6 lb of cookie dough is made from chocolate of 8 Oz

1 lb will be made by 8/ 1.6 = 0.5 Oz of chocolate

Therefore

7lb of cookie dough will be made from 7 * 0.5= 3.5 Oz of chocolate/

6 0

3 years ago

How many solutions does the equation 3x − 7 = 4 + 6 + 4x have? Two Zero Infinitely many One

sleet_krkn [62]

Answer:

one solution

Step-by-step explanation:

3x − 7 = 4 + 6 + 4x

Combine like terms

3x-7=4x+10

Subtract 3x from each side

-7=x+10

Subtract 10 from each side

-17 =x

There is one solution

4 0

3 years ago

4 Find the perimeter of a Square with a side length of 7 meters.

11Alexandr11 [23.1K]

Answer: P=28 meters

Step-by-step explanation:

$P=4$ × a ⇒ a is the side length

$P=4$ × $7$

$P=28$

5 0

2 years ago

Read 2 more answers

How do you illustrate quadratic equation in one varible?

kvasek [131]

Answer:

A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable. One absolute rule is that the first constant "a" cannot be a zero.

Step-by-step explanation:

6 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 reductio ad absurdum. 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 reductio ad absurdum 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

3 years ago