Please I need help?!?!?!?!?!

Answer:

what value??????

Answer:

It would be B

Step-by-step explanation:

Answer:

2, probably

Step-by-step explanation:

Answer:

according to me the ans is 3.

Answer:

The shelf life of a particular dairy product is normally distributed with a mean of 12 days and a standard deviation of 3 days. 3. A line up for tickets to a local concert had an average (mean) waiting time of 20 minutes with a standard deviation of 4 minutes.

Step-by-step explanation:

A = 360 cm^2

Step-by-step explanation:

Using Pythagorean theorem, the length AB is given by

AB^2 = 17^2 - 8^2 = 225 cm^2

or

AB = 15 cm

Since the ratio BC: CD = 1:2 and BC = 8 cm, this means that CD = 2×(8 cm) = 16 cm

so the length BD is

BD = BC + CD = 8cm + 16 cm = 24 cm

There the area of the rectangle is

A = AB×BD

= 15cm×24cm

= 360 cm^2

Answer:

180 D

Step-by-step explanation:

Answer:

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Answer:

Let the number be x.

x doubled is 2x, and 2x is 20 greater than 200. So, we have this:

2x = 200 + 20

2x = 220

x = 110.

Answer:

The unknown number is 110.

Step-by-step explanation:

Let the unknown number be n.

Then your number, doubled, is 200 + 20 (20 greater than 200), so the pertinent equation is    2n = 200 + 20, or 2n = 220.

Then n = 110.

Check:  Is 2(110) = 200 + 20?  Is 2(110) = 220?  YES

9514 1404 393

Answer:

  11

Step-by-step explanation:

The diagonals of an isosceles trapezoid are congruent.

  DF = EG = EH +HG = 8 +3

  DF = 11

Answer:

The diagonals of an isosceles trapezoid are congruent.

 DF = EG = EH +HG = 8 +3

 DF = 11

Step-by-step explanation:

11

Answer:

x = 140

Step-by-step explanation:

The secants exterior angle theorem states that when two secants (lines that intersect a circle at two points) intersect each other outside of the circle, then the absolute value of the difference divided by (2) equals the angle formed by the intersection of the two secants. One can apply this theorem here by stating the following, call the measure of the unknown arc (y);

[tex]\frac{30-y}{2}=10[/tex]

Solve for (y) with inverse operations:

[tex]\frac{30-y}{2}=10\\\\30-y=20\\\\y = 10[/tex]

When there is a diameter in a circle, the degree measure of the arc surrounding the diameter is (180). One can apply this here by stating the following:

[tex]30 + x + 10 = 180[/tex]

Simplify,

[tex]40 + x = 180[/tex]

Inverse operations,

[tex]40+x=180\\\\x = 140[/tex]

Answer:

y = (x+8)(x+3)

Step-by-step explanation:

The equivalent expression to the given quadratic equation is:

y = (x + 8)*(x + 3).

Here we start with:

[tex]y = x^2 + 11x + 24[/tex]

To solve the problem, we need to find the roots of the quadratic equation, to do it, we will use the Bhaskara's formula:

[tex]x = \frac{-11 \pm \sqrt{11^2 - 4*24} }{2} \\\\x = \frac{-11 \pm 5 }{2}[/tex]

Then the two solutions are:

Meaning that we can write the cuadratic equation as:

y = (x + 3)*(x + 8).

So the correct option is the first one.

If you want to learn more about quadratic equations:

https://brainly.com/question/1214333

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Hi there!

[tex]\large\boxed{AC \approx 483 ft}}[/tex]

AC is the hypotenuse, so we can use a trig formula to solve.

We are given the adjacent side, AB, so we must use cosine. Recall that:

cosθ = A/H

Thus:

cos(21.3) = 450 / H

Rearrange:

H = 450 / cos(21.3)

Use a calculator to evaluate:

H = 482.99 ≈ 483 ft.

Answer:

AC ≈ 483

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos21.3° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{AB}{AC}[/tex] = [tex]\frac{450}{AC}[/tex] ( multiply both sides by AC )

AC × cos21.3° = 450 ( divide both sides by cos21.3° )

AC = [tex]\frac{450}{cos21.3}[/tex] ≈ 483 ( to the nearest whole number )

Answer:

x=2.86 which rounds up to 3 if u need to round

Step-by-step explanation:

Answer:

why are you running

Step-by-step explanation:

why are you running

Answer:

12√3

Step-by-step explanation:

sin 60° = 18/h

h = 18/sin 60°

h = 12√3

Answer:

1,500

Step-by-step explanation:

a + 5d = 39 (1)

a + 18d = 78 (2)

Subtract (1) from (2) to eliminate a

18d - 5d = 78 - 39

13d = 39

d = 39/13

d = 3

Substitute d = 3 into (1)

a + 5d = 39 (1)

a + 5(3) = 39

a + 15 = 39

a = 39 - 15

a = 24

Sum of the first 25 terms

Sn = n/2[2a + (n – 1)d]

S25 = 25/2{2*24 + (25-1)3}

= 12.5{48 + (24)3}

= 12.5{48 + 72)

= 600 + 900

= 1,500

S25 = 1,500

Answer:did u ever get it

Step-by-step explanation:

Answer:

The answer to the problem is 15.

Answer:

Step-by-step explanation:

take 30 degree as reference angle

using cos rule

cos 30=adjacent/hypotenuse

[tex]\sqrt{3}[/tex]/2=12/y(do cross multiplication)

12*2=y[tex]\sqrt{3}[/tex]

24/[tex]\sqrt{3}[/tex]=y

8[tex]\sqrt{3}[/tex]=y

for x

using pythagors theorem

H^2=P^2+B^2

24^2=X^2+(8[tex]\sqrt{3}[/tex] )^2

576=X^2+192

576-192=X^2

384=x^2

[tex]\sqrt{384}[/tex]=x

8[tex]\sqrt{6}[/tex]=x

Answer:

48°

Step-by-step explanation:

The angle CRS looks like a "L" shape, meaning that both lines are perpindicular to each other, resulting in a right angle (which is 90°)

90° + 42° = 132°

180° - 132° = 48°

Answer:

<RCS = 48 degrees

Step-by-step explanation:

I'm pretty sure that is a right triangle

180-90-42=48

Answer:

n geometry, the notion of a connection makes precise the idea of transporting data[further explanation needed] along a curve or family of curves in a parallel and consistent manner. There are various kinds of connections in modern geometry, depending on what sort of data one wants to transport. For instance, an affine connection, the most elementary type of connection, gives a means for parallel transport of tangent vectors on a manifold from one point to another along a curve. An affine connection is typically given in the form of a covariant derivative, which gives a means for taking directional derivatives of vector fields, measuring the deviation of a vector field from being parallel in a given direction.

Connections are of central importance in modern geometry in large part because they allow a comparison between the local geometry at one point and the local geometry at another point. Differential geometry embraces several variations on the connection theme, which fall into two major groups: the infinitesimal and the local theory. The local theory concerns itself primarily with notions of parallel transport and holonomy. The infinitesimal theory concerns itself with the differentiation of geometric data. Thus a covariant derivative is a way of specifying a derivative of a vector field along another vector field on a manifold. A Cartan connection is a way of formulating some aspects of connection theory using differential forms and Lie groups. An Ehresmann connection is a connection in a fibre bundle or a principal bundle by specifying the allowed directions of motion of the field. A Koszul connection is a connection which defines directional derivative for sections of a vector bundle more general than the tangent bundle.

Connections also lead to convenient formulations of geometric invariants, such as the curvature (see also curvature tensor and curvature form), and torsion tensor.

Step-by-step explanation:

Answer:

n geometry, the notion of a connection makes precise the idea of transporting data[further explanation needed] along a curve or family of curves in a parallel and consistent manner. There are various kinds of connections in modern geometry, depending on what sort of data one wants to transport. For instance, an affine connection, the most elementary type of connection, gives a means for parallel transport of tangent vectors on a manifold from one point to another along a curve. An affine connection is typically given in the form of a covariant derivative, which gives a means for taking directional derivatives of vector fields, measuring the deviation of a vector field from being parallel in a given direction.

Connections are of central importance in modern geometry in large part because they allow a comparison between the local geometry at one point and the local geometry at another point. Differential geometry embraces several variations on the connection theme, which fall into two major groups: the infinitesimal and the local theory. The local theory concerns itself primarily with notions of parallel transport and holonomy. The infinitesimal theory concerns itself with the differentiation of geometric data. Thus a covariant derivative is a way of specifying a derivative of a vector field along another vector field on a manifold. A Cartan connection is a way of formulating some aspects of connection theory using differential forms and Lie groups. An Ehresmann connection is a connection in a fibre bundle or a principal bundle by specifying the allowed directions of motion of the field. A Koszul connection is a connection which defines directional derivative for sections of a vector bundle more general than the tangent bundle.

Connections also lead to convenient formulations of geometric invariants, such as the curvature (see also curvature tensor and curvature form), and torsion tensor.

Step-by-step explanation:

Answer:

yes, your right!

Step-by-step explanation:

yes you're right and this you done by using Pythagoras theorem!

Answer:

60

Step-by-step explanation:

Answer:

60°

Step-by-step explanation:

Answer:

10x+ y

Step-by-step explanation:

The unit's digit is y and the ten's digit is x.

The ten's digit has a zero placed beside it .

So multiply x by 10  giving 10 x and then add the unit's digit .

This wil give 10x+ y

The number is 10 x + y

This can be elaborated through the use of numbers . Suppose we have unit's digit as 6 and the ten's digit as 5.

Multiply 10 by 5 and add 6

5*10 +6= 50+6= 56

Step-by-step explanation:

it should be x= -3 and x= 1

Answer:

15

Step-by-step explanation:

To find the unit rate you divide.

• the points that is in bold is ( 2, 30) from that information you would see what times 2 gives me 30.

• 30/2 =15

• unit rate= 15

Answer:

slope=y2-y1/x2-x1

=5-5/-1-3

=0/-4

=0

Step-by-step explanation:

Answer:

slope = 0

Step-by-step explanation:

Calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (3, 5) and (x₂, y₂ ) = (- 1, 5)

m = [tex]\frac{5-5}{-1-3}[/tex] = [tex]\frac{0}{-4}[/tex] = 0