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2 years ago

The area of square abcd is 7cm^2 show that x^2+4x=3

Mathematics

1 answer:

AveGali [126]2 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

The area (A) of the square = s² ( s is the measure of the side )

Here s = x + 2 and A = 7 , thus

(x + 2)² = 7 ← expand left side using FOIL

x² + 4x + 4 = 7 ( subtract 4 from both sides )

x² + 4x = 3 ← as required

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Rounded 6,289,002 to the nearest 1,000,000

kotegsom [21]

<span>6,289,002 rounded to the nearest 1,000,000 is 6,000,000. This is because the number in the hundred thousands column, the one to the right of the first digit, is less than five, so it gets rounded down.</span>

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2 years ago

Steve determined the number of states each of his friends has visited. A number line goes from 0 to 44. The whiskers range from

Mnenie [13.5K]

Answer:

Step-by-step explanation:

I got it right on edge

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2 years ago

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Squaring both sides of an equation is irreversible. Is Cubing both sides of an equation reversible? Provide numerical examples t

nikitadnepr [17]

Answer:

<h2>Cubing both sides of an equation is reversible.</h2>

Step-by-step explanation:

Squaring both sides of an equation is irreversible, because the square power of negative number gives a positive result, but you can't have a negative base with a positive number, given that the square root of a negative number doesn't exist for real numbers.

In case of cubic powers, this action is reversible, because the cubic root of a negative number is also a negative number. For example

$\sqrt[3]{x} =-1$

We cube both sides

$(\sqrt[3]{x} )^{3} =(-1)^{3} \\x=-1$

If we want to reverse the equation to the beginning, we can do it, using a cubic root on each side

$\sqrt[3]{x}=\sqrt[3]{-1} \\\sqrt[3]{x}=-1$

There you have it, cubing both sides of an equation is reversible.

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1 year ago

Sarah and thomas are going bowling. the probability that sarah scores more than 175 is 0.4, and the probability that thomas scor

pychu [463]

When the occurrence of one event say A does not affect the occurrence of another event say B, than the two events are said to be independent such that;

$\\ 
P(A\cap B)=P(A)\times P(B)$

where, P(A) = probability of occurrence of event A

and P(B) = probability of occurrence of event B

(a).

Now, let event A = Sarah scores more than 175

and event B = Thomas scores more than 175

Thus, P(A)= Probability that Sarah scores more than 175 = 0.4

and P(B)= Probability that Thomas scores more than 175 = 0.2

Since, the scores are independent, thus the probability that both Sarah and Thomas scores more than 175 is,

$\\ 
P(A\cap B)=P(A)\times P(B)\\ 
P(A\cap B)= 0.4\times 0.2= 0.08\\$

Hence, the required probability is 0.08

(b).

When the occurrence of one event say A affects the occurrence of another event say B, than the two events are said to be dependent such that;

$\\ 
P(A\cap B)=P(A)\times P(B\setminus A)\\$

Now, let event A = Sarah scores more than 175

and event B = Thomas scores more than 175

Thus, P(A)= Probability that Sarah scores more than 175 = 0.4

P(B)= Probability that Thomas scores more than 175 = 0.2

and P(B|A) = Sarah scores more than Thomas given that Thomas scores more than 175 = 0.3

Thus, the required probability is calculated as follows;

$\\ 
P(A\cap B)=P(A)\times P(B\setminus A)\\ 
P(A\cap B)=0.2\times 0.3=0.06$

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2 years ago

QUICK! 75 POINTS !!Select all that are part of the solution set of csc(x) > 1 and over 0 ≤ x ≤ 2π.

Vladimir79 [104]

Answer:

$\frac{\pi}{4}$

$\frac{5\pi}{6}$

Step-by-step explanation:

The answer uses the unit circle and that sine and cosecant are reciprocals.

The first choice doesn't even fit the criteria that $x$ is between $0$ and $2\pi$ (inclusive of both endpoints) because of the $x=\frac{-7\pi}{6}$ .

Let's check the second choice.

$\csc(\frac{\pi}{4})=\frac{2}{\sqrt{2}} \text{ since } \sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ .

$\csc(\frac{\pi}{4})>1 \text{ since } \frac{2}{\sqrt{2}}>1$

$\csc(\frac{\pi}{2})=1 \text{ since } \sin(\frac{\pi}{2})=1$ which means $\csc(\frac{\pi}{2})=1$ which is not greater than 1.

So we can eliminate second choice.

Let's look at the third.

$\csc(\frac{5\pi}{6})=2 \text{ since } \sin(\frac{5\pi}{6})=\frac{1}{2}$ which means $\csc(\frac{5\pi}{6})>1$ .

$\csc(\pi)$ isn't defined because $\sin(\pi)=0$ .

So we are eliminating 3rd choice now.

Let's look at the fourth choice.

$\csc(\frac{7\pi}{6})=-2 \text{ since } \sin(\frac{7\pi}{6})=\frac{-1}{2}$ which means $\csc(\frac{7\pi}{6})$ and not greater than 1.

I was looking at the rows as if they were choices.

Let me break up my choices.

So we said $x=-\frac{7\pi}{6}$ doesn't work because it is not included in the inequality $0\le x \le 2\pi$ .

How about $x=0$ ? This leads to $\csc(0)$ which doesn't exist because $\sin(0)=0$ .

So neither of the first two choices on the first row.

Let's look at the second row again.

We said $\frac{\pi}{4}$ worked but not $\frac{\pi}{2}$

Let's look at the choices on the third row.

We said $\frac{5\pi}{6}$ worked but not $x=\pi$

Let's look at at the last choice.

We said it gave something less than 1 so this choice doesn't work.

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2 years ago

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