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3 years ago

(10 POINTS AND BRAINLIEST FOR BEST)

Mathematics

2 answers:

balandron [24]3 years ago

8 0

Answer:

B. 0.85(96.82)

Step-by-step explanation:

Let the total cost of items purchased before discount = 1

Discount on purchases = 15 % = 0.15

Cost of items purchased after discount = (1 - discount) × cost of items in tiffany's cart

Cost of items purchased after discount = (1 - 0.15) of 96.82

Cost of item purchased after discount = 0.85 (96.82)

Therefore option B is the correct answer.

kherson [118]3 years ago

4 0

I think its A im not exactly sure it also could be D.

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WRITE INEQUALITIE TO DESCRIBE A REGION CONSISTING OF ALL POINTS BETWEEN ( BUT NOT ON ) THE SPHERES OF RADIUS R AND r CENTERED AT

Andreas93 [3]

Answer:

$r^2$

Step-by-step explanation:

We are given the following in the question:

$r < R$

where r and r are the radius of the circle.

General equation of circle:

$(x-h)^2 + (y-k)^2 + (z-j)^2 = r^2$

where(h,k,j) is the center of the circle and r is the radius of circle.

If the circle is centered at origin, then,

$x^2 + y^2+z^2 = r^2$

Equation of circle with radius R centered on origin

$x^2 + y^2+z^2 = R^2$

Inequality to describe the region that consist of all points lying between the sphere of radius r and R but not on the sphere is given by:

$r^2$

4 0

3 years ago

Prove Sin3theta = 3sintheta - 4sin^3theta

trapecia [35]

Express the left hand side as

sin3theta=sin(theta+2theta)

now expand the right side of this equation using color(blue)"Addition formula"

color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))

rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))

The right hand side is expressed only in terms of sintheta's

so we use cos2theta=1-2sin^2theta........(1)

color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)

Replace cos2theta" and " sin2theta by the expansions (1) and (2)
into (A)

sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)

and expanding brackets gives.

sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))

Replace cos^2theta=1-sin^2theta" into (B)"

rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)

and expanding 2nd bracket gives.

sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta

Finally, collecting like terms.

sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"

3 0

3 years ago

Find the distance between the points given.

Y_Kistochka [10]

So the answer is (C)

c.2/5

hope this HELP

3 0

3 years ago

What is the length of segment AC

lakkis [162]

To find the length of segment AC, we must find the total rise and total run between the two points.

Point C is located at (-5, 5). Point A is located at (3,-1). To find the rise, subtract the y-value of A from the y-value of C:

$5 - (-1) = 6$

The rise of this segment is 6.

To find the run, subtract the x-value of A from the x-value of C:

$3 - (-5) = 8$

The run of this segment is 8.

We can use the Pythagorean Theorem to find the length of this segment. The theorem uses the following formula:

$a^{2} + b^{2} = c^{2}$

The segment represents the hypotenuse, and the rise and run represent the legs of this segment. We know that the two legs' lengths are 6 and 8, so plug them into the equation:

$6^{2} + 8^{2} = c^{2}$
$36 + 64 = c^{2}$
$100 = c^{2}$

Square root both sides to get c by itself:

$\sqrt{100} = 10$
$c = 10$

The length of segment AC is 10.

6 0

4 years ago

Read 2 more answers

Find the distance between the points (4, –2) and (0, 10).

madam [21]

$A(4,-2) \\B(0, 10) \\AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\AB=\sqrt{(0-4)^2+(10-(-2))^2} \\AB=\sqrt{16+144} \\AB=\sqrt{160}\approx\boxed{12.65} \\$

The answer is D.

Hope this helps.

r3t40

8 0

4 years ago