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3 years ago

How do I express four less than the square number

Mathematics

2 answers:

netineya [11]3 years ago

5 0

Answer:

x^2 - 4

Step-by-step explanation:

Number squared (x)^2

less than is a subtraction term

put them together you get x^2 - 4

kobusy [5.1K]3 years ago

4 0

Answer:

X^2-4

Step-by-step explanation:

Let x be our number.

So we square the number because it says 4 less than the square number.

So we have x^2.

Now, we have to take 4 less than that.

So x^2-4 is our answer.

(This can also be factored because it is in the form a^2-b^2)

(This can be rewritten as (x+2)(x-2))

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3 years ago

Find the slope of the line. A) -3/4 B) 3/4 C) 4/3 D) -4/3

Slav-nsk [51]

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3 years ago

Read 2 more answers

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

4 years ago

If x > 0 and y > 0, where is the point (x, y) located?

Scrat [10]

Given, x > 0, y > 0.
So, the values of x and y are less than 0, that means, they are negative integers.
If we divide a Cartesian plane with the x-axis and y-axis, we get four quadrants.
1st Quadrant : (+,+)
2nd Quadrant : (-,+)
3rd Quadrant : (-,-)
4th Quadrant : (+,-)
Since the values of x and y are both negative, so (x, y) is located in the 3rd Quadrant.

Hope you could get an idea from here.

Doubt clarification - use comment section

8 0

2 years ago