HELP ASAP ITS TIMED
Which equation is related to √x+10-1=x?
O x + 10 = x2 + x + 1
O x + 10 = x2 + 2x + 1
O x + 10 = x² +1
Ox+10=x²-1

Step-by-step explanation:

Greetings !

Firstly, find the class-midpoints (X) to solve for the mean.

Thus, mid points of each classes are

\(5.15.25.35.45\)

Solve, for p using mean grouped data formula

\( \frac{5(3) + 15(4) + 25(p) + 35(3) + 45(2)}{3 + 4 + p + 3 + 2} = 24\)

multiply each values add them and solve for p.

\( \frac{15 + 60 + 25p + 105 + 90}{12 + p} = 24\)

make a cross multiplication

\(270 + 25p = 24(12 + p) \\ 270 + 25p = 288 + 24p\)

subtract 270 from both sides

\(270 + 25p - 270 = 288 + 24p - 270 \\ 25p = 24p + 18\)

subtract 24p from both sides

\(25p - 24p = 24p + 18 - 24p \\ p = 18\)

Also refer the attached image for further details

Hope it helps!

The length of the longer ship in terms of number of football fields is;

4 football fields

Let the length of ship 1 be x

Let the length of ship 2 be y

Total length of both ships is 2321 ft

Thus;

x + y = 2321  ----(eq 1)

The difference in their lengths is 17 ft

Thus;

x - y = 17    ----(eq 2)

Add eq 1 to eq 2 to get;

2x = 2338

Thus; x = 2338/2

x = 1169 ft

Put 1169 for x in eq 1 to get;

1169 + y = 2321

y = 2321 - 1169

y = 1152 ft

Longer ship = 1169 ft long

From conversion tables;

1 ft = 1/3 yards

Thus; 1169 ft = 1169 * 1/3 = 389.67 yards

A football field has a length of 100 yards

Then number of football fields for the length of the longer ship = 389.67/100 = 3.8967 ≈ 4 football fields.

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Ceva's Theorem states that given any triangle △ABC, the segments from A, B, and C to the opposite sides of the triangle are concurrent precisely when the product of the ratios of the pairs of segments formed on each side of the triangle is equal to 1.

The distance between the tops of the buildings is 28.5 meters.

To find the distance between the top of the buildings, we can use the concept of similar triangles.

Let's denote the height of the shorter building as "a" (12 m) and the height of the taller building as "b" (19 m). The distance between the buildings can be denoted as "c" (18 m), and the distance between the top of the buildings as "x" (which we need to find).

We can set up a proportion based on the similar triangles formed by the buildings:

a/c = b/x

Substituting the known values:

12/18 = 19/x

To find "x," we can cross-multiply and solve for "x":

12x = 18 * 19

12x = 342

x = 342/12

x = 28.5 m

Therefore, the distance between the tops of the buildings is 28.5 meters.

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Solution:

(g o g)(x) = -((-x^2 + 5)^2) + 5

Answer:

(g o g)(x) = -x^4 + 10x^2 - 20

Answer:

rad 29

Step-by-step explanation:

diagonal is 16+4=20 so \(\sqrt{20}\)

3 and rad20 you will have 20 +9=29

answer is \(\sqrt{29}\)

Answer:

B. 37 units

Step-by-step explanation:

DE = 4x + 1

AB = 11x - 25

DE is a midsegment of the ∆ABC

Therefore:

DE = ½(AB) => Triangle Midsegment Theorem

4x + 1 = ½(11x - 25)

2(4x + 1) = 11x - 25

8x + 2 = 11x - 25

Collect like terms

8x - 11x = -2 - 25

-3x = -27

-3x/-3 = -27/-3

x = 9

✔️DE = 4x + 1

Plug in the value of x

DE = 4(9) + 1 = 36 + 1

DE = 37

y=9

x=12

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Answer:

Step-by-step explanation:

Well, since the length of a month can vary in the number of days, this answer can also vary.

For example, February is only 28 days long, while December is 31 days long.

That being said, the average length of all 12 months is 30.436875 days, so if Denny left to another country on June 20th, he would most likely be back   July 19th of July 20th.

I hope this helps!

Answer:

(-8,1)

Step-by-step explanation:

Imma assume you meant (x-2, y)

x coordinate becomes -6-2=-8

y coordinate stays the same

so the image is (-8,1)

Answer:

  x < -1

Step-by-step explanation:

Since the parabola opens upward, it is positive and decreasing where the left branch is above the x-axis: all points to the left of x=-1.

  all real values of x where x < -1

Answer:

A positive number or a number that is greater than zero

Step-by-step explanation:

x+2

Step-by-step explanation:

x+2

If x = –2 is a zero, then x + 2 = 0, so x + 2 is a factor. Similarly, if x = 1/3 is a zero, then x – 1/3 = 0, so x – 1/3 is a factor. By giving me two of the zeroes, they have also given me two factors: x + 2 and x – 1/3.

Binomial expansion is used to factor expressions that can be expressed as the power of the sum of two numbers.

The proof that (1.01)^12 exceeds (1.02)^6 by 0.0007 is\(\mathbf{(1.01)^{12} - (1.02)^6 \approx 0.0007 }\)

The expressions are given as:

\(\mathbf{(1.01)^{12}\ and\ (1.02)^6}\)

A binomial expression is represented as:

\(\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}\)

Express 1.01 as 1 + 0.01

So, we have:

\(\mathbf{(1.01)^{12} = (1 + 0.01)^{12}}\)

Apply the above formula

\(\mathbf{(1.01)^{12} = ^{12}C_0 \times 1^{12 - 0} \times 0.01^0 + ......... .......... + ^{12}C_{12} \times 1^{12 - 12} \times 0.01^{12} }}\)

\(\mathbf{(1.01)^{12} = 1 \times 1 \times 1 + ......... .......... + 1 \times 1 \times 10^{-24} }}\)

\(\mathbf{(1.01)^{12} = 1 + ......... .......... + 10^{-24} }}\)

This gives

\(\mathbf{(1.01)^{12} = 1.1268\ (approximated)}\)

Similarly,

Express 1.02 as 1 + 0.02

So, we have:

\(\mathbf{(1.02)^6 = (1 + 0.02)^6}\)

Apply \(\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}\)

\(\mathbf{(1.02)^6 = ^6C_0 \times 1^{6 - 0} \times 0.02^0 + ^6C_1 \times 1^{6 - 1} \times 0.02^1 +.............. + ^6C_6 \times 1^{6 - 6} \times 0.02^6 }\)\(\mathbf{(1.02)^6 = 1 \times 1 \times 1 + 6 \times 1 \times 0.02 +.............. + 1 \times 1 \times 6.4 \times 10^{-11} }\)

\(\mathbf{(1.02)^6 = 1 + 0.12 +.............. + 6.4 \times 10^{-11} }\)

This gives

\(\mathbf{(1.02)^6 = 1.1261\ (approximated) }\)

Calculate the difference as follows:

\(\mathbf{(1.01)^{12} - (1.02)^6 \approx 1.1268 - 1.1261 }\)

\(\mathbf{(1.01)^{12} - (1.02)^6 \approx 0.0007 }\)

The above equation means that:

(1.01)^12 exceed the value of (1.02)^6 by 0.0007

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Answer: a = 10

radius = 7

Step-by-step explanation:

The generic equation for a circle of radius R is, centered in the point (a, b)

(x - a)^2 + (x - b)^2 = R^2

In this case, we know that the center is (3, -5)

Then the equation of the circle must be something like:

(x - 3)^2 + (y + 5)^2 = R^2

And we have:

x^2 + y^2 - 6*x + a*y = 15

First, let's complete squares for x:

(x - 3)^2 = x^2  + 2*(-3)*x + (-3)^2 = x^2 - 6*x + 9.

Then we can add 9 to both sides of the equation to get:

x^2 + y^2 - 6*x + a*y + 9 = 15 + 9

(x^2 - 6*x + 9) + y^2 + a*y = 24

(x - 3)^2 + y^2 + a*y = 24

Now let's do the same for y:

(y + 5)^2 = y^2 + 2*5*y + 5^2 = y^2 + 10*y + 25

From this we can already see that a must be equal to 10, if we replace a by 10, the circle equation becomes:

(x - 3)^2 + y^2 + 10*y = 24

Now we can add 25 to both sides of the equation to get:

(x - 3)^2 + y^2 + 10*y + 25 = 24 + 25

(x - 3)^2 + (y + 5)^2 = 49

An the radius of the circle will be equal to:

r = √49 = 7

Answer:

it is an acute scalene triangle

Answer:

Since the lengths of the triangle are 4,5, and 6, which are almost equal, then the triangle cannot be obtuse. It's right triangle

Answer:

3/ 25

Step-by-step explanation:

3 squares

Each square inside those is equal to 25 cubes.

3/ 25

Answer: (0,10)

Step-by-step explanation: y=mx+b

y=12.5x+10

0=12.5x+10

10 is y because its on the outside

Answer:

229.41

Step-by-step explanation:

Well you need to multiply 11.3 times 4.5 per each triangle

Also you need to find the base which is a square

So 5.1 times 5.1 which is technically 5.1 squared. You add from there

EQUATION: 4(11.3 x 4.5) + 5.1 squared  

You should get 229.41

Answer:

7(1 + 6)

Step-by-step explanation:

Answer:

Four thousand two hundred and

Step-by-step explanation:

The distance from the north most ship to the observation is 2.9 miles and the distance from the south most ship to the observation is 9.9 miles.

The bearing is defined as the angle measured clockwise from the north direction. A ship is anchored off a long straight shoreline that runs north and south.

From two observation points miles apart on shore, the bearings of the ship are 52 degrees and 134 degrees respectively. To find the distance from the ship to each of the observation points, we can use trigonometry.

Let's call the distance from the north observation point to the ship x, and the distance from the south observation point to the ship y. The bearings from the north and south observation points can be drawn as follows:

[asy]

unitsize(1cm);

pair O = (0,0), N = (-2,0), S = (2,0), A = (-2,1), B = (2,-1);

draw(O--N--A,Arrow);

draw(O--S--B,Arrow);

\(label("$52^\circ$", N + 0.3*dir(52), NE);\)

\(label("$134^\circ$", S + 0.3*dir(134), SE);\)

draw(O--A--B--cycle,dashed);

\(label("$x$", (N+A)/2, W);\)

\(label("$y$", (S+B)/2, E);\)

[/asy]

We can use the tangent function to find x and y, since we have the angle and opposite side.

From the north observation point, we have:

\($$\tan(52^\circ) = \frac{x}{2}$$$$x = 2\tan(52^\circ) \approx 2.95$$\)

From the south observation point, we have:

\($$\tan(134^\circ) = \frac{y}{2}$$$$y = 2\tan(134^\circ) \approx 9.88$$\)

Therefore, the distance from the north most ship to the observation is 2.9 miles and the distance from the south most ship to the observation is 9.9 miles.

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Answer: the answer is well i cant see anything so your going to have to find out by yourself

Step-by-step explanation: Im sorry

Answer:

Step-by-step explanation:

m∠QDC = 5*12 + 17 = 77

Anwser:

I think it is 82 because it goes down 2 than sunday it goes up 2 to put it back at 82.

Answer:

Step-by-step explanation:

9. sin 18 = x/21

 x = 21* sin 18

  x = 6.5

10. tan 55 = 29/x

  x* tan 55 = 29

x = 29/tan 55

x = 20.3

11. sin 23 = x/25

  x = 25 * sin 23

  x = 9.8

12. sin 47 = 33/x

  x * sin 47 = 33

 x = 33/sin 47

  x = 45.1

Answer:

\(\huge\boxed{z=6;\ x=12.5}\)

Step-by-step explanation:

If two figures are similar, then corresponding sides are in proportion.

Therefore:

\(\dfrac{z}{15}=\dfrac{4}{10}\qquad|\text{cross multiply}\\\\10z=(15)(4)\\\\10z=60\qquad|\text{divide both sides by 10}\\\\\boxed{z=6}\)

\(\dfrac{x}{5}=\dfrac{10}{4}\qquad|\text{cross multiply}\\\\4x=(5)(10)\\\\4x=50\qquad|\text{divide both sides by 4}\\\\x=\dfrac{50}{4}\\\\\boxed{x=12.5}\)