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3 years ago

14

Explain the amendment process in your own words. Or, you can use the following chart to fill in details about the amendment proc

ess

1 answer:

crimeas [40]3 years ago

8 0

Answer:

i can answer if you tell me which amendment????

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago

Help math ITS EASY IF U LOKE RATIO

scoundrel [369]

You have the right one selected.

The answer is 2:1

To find this take each side of ABC and its corresponding side on DEF, then simplify. You can treat ratios just like fractions, so simplify them in the same way.

20:10 becomes 2:1

12:6 becomes 2:1

16:8 becomes 2:1

7 0

3 years ago

Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers

How many real zeros does the function graph below have?

Nostrana [21]

Answer:

Step-by-step explanation:

If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity.

6 0

3 years ago

I don’t get. Can I aleast get someone to explain how to do it

Nastasia [14]

You have to make the percent a decimal by moving the decimal to the left twice. then you need to multiply the other number.

5 0

3 years ago

Read 2 more answers