What is the value of x? a) 18° b) 36° c) 43° d) 86°

Mathematics

1 answer:

dmitriy555 [2]2 years ago

7 0

Answer:

I think is 18

Step-by-step explanation:

because is the same of the L

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Points q, r, s, t, are plotted on the number line below

vodomira [7]

Answer:

Step-by-step explanation:

must be r and s

because right at in middle of the distance between them is 0

4 0

2 years ago

What is 1/6d+2/3 = 1/4(d-2)

GenaCL600 [577]

1/6d + 2/3 = 1/4(d - 2)

First, simplify $\frac{1}{6} d$ to $\frac{d}{6}$ / Your problem should look like: $\frac{d}{6}$ + $\frac{2}{3}$ = $\frac{1}{4}$ (d - 2)
Second, simplify $\frac{1}{4} (d - 2)$ to $\frac{d-2}{4}$ / Your problem should look like: $\frac{d}{6}$ + $\frac{2}{3}$ = $\frac{d-2}{4}$
Third, multiply both sides by 12 (the LCM of 6,4) / Your problem should look like: 2d + 8 = 3(d - 2)
Fourth, expand. / Your problem should look like: 2d + 8 = 3d - 6
Fifth, subtract 2d from both sides. / Your problem should look like: 8 = 3d - 6 - 2d
Sixth, simplify 3d - 6 - 2d to d - 6 / Your problem should look like: 8 = d - 6
Seventh,add 6 to both sides. / Your problem should look like: 8 + 6 = d
Eighth, simplify 8 + 6 to 14 / Your problem should look like:14 = d
Ninth, switch sides. / Your problem should look like: d = 14

Answer: d = 14

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3 years ago

Question 12 (1 point)

Paraphin [41]

<h3>Answer: 5</h3>

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Explanation:

Let's consider the expression (x-y)^2. It expands out to x^2-2xy+y^2. The terms are:

x^2
-2xy
y^2

Each of those terms either has a single variable with an exponent of 2, or has the exponents add to 2. Think of 2xy as 2x^1y^1.

In short, this means that the degree of each monomial term is 2.

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Now consider (x-y)^3. It expands out into x^3-3x^2y+3xy^2+y^3.

We have terms that either have a single variable and the exponent is 3, or the exponents add to 3. The degree of each term is 3.

----------

This pattern continues.

In general, for (x-y)^n, where n is any positive whole number, the degree of each term in the expansion is n. If you picked any term, added the exponents, then the exponents will add to n.

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2 years ago

ABC has the points A(1,7) B(-2,2) and C(4,2) as it’s vertices

tekilochka [14]

Part 1

Using the distance formula,

$AB=\sqrt{(-2-1)^{2}+(2-7)^{2}}=\sqrt{34}\\\\BC=\sqrt{(-2-4)^{2}+(2-2)^{2}}=6\\\\AC=\sqrt{(1-4)^{2}+(7-2)^{2}}=\sqrt{34}$