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DerKrebs [107]
3 years ago
8

What’s the answer to 10(1-x)

Mathematics
1 answer:
NeTakaya3 years ago
5 0

Answer:

10-10x

Step-by-step explanation:

All you do is distribute the 10. You can't add numbers if they don't both have the same variable, so you leave it as 10-10x.

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You have just arranged for a $1,640,000 mortgage to finance the purchase of a large tract of land. the mortgage has an apr of 6.
sergeinik [125]
You would take 6.4% into 1640000 and you would pay that the next 30 years

3 0
3 years ago
Help! How would I solve this trig identity?
NeTakaya

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}

Then the identity can be rewritten as:

sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}  = \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)} \\\\

Now we can multiply both sides by cos⁴(x) to get:

\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}  = \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)}  + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)

Now we can use the identity:

sin²(x) + cos²(x) = 1

1 - cos^2(x) = sin^2(x)*(sin^2(x) + cos^2(x)) = sin^2(x)\\\\1 = sin^2(x) + cos^2(x) = 1

Thus, the identity was proven.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

#SPJ1

7 0
1 year ago
Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane
postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

(c) (x - 3)² + (y + 4)² + (z - 5)² = 16

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r²            ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i.  Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}

d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}

d = \sqrt{(0)^2+ (0)^2 + (-5)^2}

d = \sqrt{(25)}

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²  

(x - 3)² + (y + 4)² + (z - 5)² = 25  

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25  

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i.  Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}

d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}

d = \sqrt{(-3)^2 + (0)^2+ (0)^2}

d = \sqrt{(9)}

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²  

(x - 3)² + (y + 4)² + (z - 5)² = 9  

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9  

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i.  Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}

d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}

d = \sqrt{(0)^2 + (4)^2+ (0)^2}

d = \sqrt{(16)}

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²  

(x - 3)² + (y + 4)² + (z - 5)² = 16  

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

 

3 0
3 years ago
372-199= What’s the answer
Montano1993 [528]

Answer:

Step-by-step explanation:

The answer is 173. Just use the calculator

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