Determine
the rotation around the origin that was performed to transform the preimage on the left to its image on the right.

Answer:

7 inches

Step-by-step explanation:

The surface area of a cube is denoted by: A = 6s², where s is the side length.

Here, we know that A = 294, so plug this into the formula to solve for s:

A = 6s²

294 = 6s²

s² = 49

s = √49 = 7

Thus, the answer is 7 inches.

I know you're looking for an answer but i'm doing a challenge where i have to answer 10 questions by the end of today, I'm sorry! :(

Using the binomial distribution, it is found that there is a 0.999977 = 99.9977% probability that at least one of them makes a Type I error.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

In this problem:

The probability that at least one commits an error is given by:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{30,0}.(0.3)^{0}.(0.7)^{30} = 0.000023[/tex]

Then:

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.000023 = 0.999977[/tex]

There is a 0.999977 = 99.9977% probability that at least one of them makes a Type I error.

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Answer:

1200

Step-by-step explanation:

since the first week has 75 doubling it is 150 doubling that is 300 doubling that is 600 doubling that for the final week is 1200.

1200 more zombies will there be after an additional 4 weeks.

A sequence is an enumerated group of items in mathematics where repetitions are permitted and order is important. It has members (also known as elements or phrases), just like a set. The length of the series is the number of elements (potentially infinite).

At initial the zombies are= 75

Now, the zombies doubles at every week.

So, After 1 st week the population of zombies are

= 75 x 2

= 150

Then, After 2 nd week the population of zombies are

= 150 x 2

= 300

Then, After 3 rd week the population of zombies are

= 300 x 2

= 600

Then, After 4th week the population of zombies are

= 300 x 2

= 1200

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Answer:

50.29in

Step-by-step instructions:

First find the semi-circular part:- 1/2×2πr

=1/2×2×22/7×9

=28.29

Find whole perimeter:- 28.29+5+5+12 = 50.29 in

Answer: 44 degrees

Step-by-step explanation:

Answer:

5/6

Step-by-step explanation:

it comes out originally like 10/12 but if your simplify it it's 5/6

Answer:5/6

Step-by-step explanation: 12-2 =10 so 10/12 which simplifies to 5/6

Answer:

She translated each vertex of Figure 1 seven units to the right.

Step-by-step explanation:

Answer: x = 5, y = 7

Step-by-step explanation:

1. First, put the variables together/rearrange:

y - x = 2 & 5x - 4y = -3

2. Cancel out variable x by multiply the first equation by 5:

5y - 5x = 10 & 5x - 4y = -3

3. Add equations:

y = 7

4. Then determine x by plugging y back into the first equation y = x + 2:

x = 5

Answer:

6.0 ft^2

Step-by-step explanation:

hope this helps

Hey so I have a question do u need to explain?

first add the ratio numbers which is 12

[tex] \frac{2}{12} \times 6000 = 1000[/tex]

then divide 1000 with 175

=5.71

5.71×2.25

=12.85 ruro

Answer:

2 semi circles put together

Step-by-step explanation:

just cut the circle in half

53x95 = 5035!

hopes this helps! ^^

Answer:

5605

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

I did this question before.

Answer:

Let a = side length of a cube

Let S = surface area of a cube

Area of a square = a²

Since a cube has 6 square sides:  S = 6a²

To make a the subject:

S = 6a²

Divide both sides by 6:

[tex]\sf \implies \dfrac{S}{6}=a^2[/tex]

Square root both sides:

[tex]\sf \implies a=\sqrt{\dfrac{S}{6}}[/tex]

(positive square root only as distance is positive)

-----------------------------------------------------------------------------------------------

[tex]\sf x=-3-\sqrt{2} \implies (x+[3+\sqrt{2}])=0[/tex]

[tex]\sf x=-3+\sqrt{2} \implies (x+[3-\sqrt{2}])=0[/tex]

Therefore,

[tex]\sf y=a(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]  for some constant a

Given the y-intercept is at (0, -5)

[tex]\sf \implies a(0+3+\sqrt{2}) (0+3-\sqrt{2})=-5[/tex]

[tex]\sf \implies a(3+\sqrt{2}) (3-\sqrt{2})=-5[/tex]

[tex]\sf \implies a(9-2)=-5[/tex]

[tex]\sf \implies 7a=-5[/tex]

[tex]\sf \implies a=-\dfrac57[/tex]

Substituting found value of a into the equation and simplifying:

[tex]\sf y=-\dfrac57(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])[/tex]

[tex]\sf \implies y=-\dfrac57(x^2+6x+7)[/tex]

[tex]\sf \implies y=-\dfrac57x^2-\dfrac{30}{7}x-5[/tex]

Answer:

25a^2b^4c^2

Step-by-step explanation:

step 1:

Remove the bracket by multiplying all the power inside the bracket with the power outside the bracket.

5^2a^2b^2x2c^2=25a^2b^4c^2

Answer:

its 54

Step-by-step explanation:

1+2=3 3x5=15 15x2=30 30+16=46 46+8=54

Answer:

the answer to your question is B

Answer:

t = 0, and t = 4/5

Step-by-step explanation:

1. Write the equation in standard form:

-75t^2+60t=0

2. Solve with the Quadratic formula:

[tex]t_{1,2} = \frac{-60 +- \sqrt{60^2 - 4(-75)}0}{2(-75)} \\\\t_{1,2} = \frac{-60+-60}{2(-75)} \\\\3. Separate the fractions:\\t_{1} = \frac{-60+60}{2(-75)} , t_{2} = \frac{-60-60}{2(-75)} \\\\4. Solve for each:\\t_{1} = \frac{0}{-150} = 0\\t_{2} = \frac{-120}{-150} = 4/5[/tex]

hope this helps!

The required equation f(10) = 13.52 mg remains.

We have given that ,

m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60

The amount of a radioactive substance remaining after t years is given by the function  

[tex]f(t)=m(0.5)^{t/h}[/tex]............ (1),

where m = initial mass and h= half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,

then from equation (1) we get,

[tex]f(10)=50(0.5)^{10/5.3}[/tex]

Therefore  the required equation  f(10) = 13.52 mg .

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Answer:

The required equation f(10) = 13.52 mg remains.

We have given that ,

m is the initial mass and h is the half-life in years. cobalt-60 has a half-life of about 5.3 years. which equation gives the mass of a 50 mg cobalt-60

What is the fromula for he amount of a radioactive substance remaining after t years?

The amount of a radioactive substance remaining after t years is given by the function  

............ (1),

where m = initial mass and h= half-life in years.

Now, for Cobalt-60, h = 5.3 years, m = 50 mg and t = 10 years,

then from equation (1) we get,

Therefore  the required equation  f(10) = 13.52 mg .

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Answer: X-axis

Step-by-step explanation: The arrow is pointing to the horizontal line on the on the graph. That horizontal line is the x-axis.

Answer:

200/7

Step-by-step explanation:

(2 meters)/(7 centimeters) or (2m/7cm)

1 meter = 100 cm

(2m/7cm)*(100cm/1m) = 200/7 (or   28.57) scale factor.

Answer:

[tex]\boxed{\sf{x=7 \quad y=2}}[/tex]

Step-by-step explanation:

To solve this problem, isolate it on one side of the equation.

x-3y=1: x=1+3y

x=1+3y

Substitute.

x=1+3y= 3(1+3y)-5y=11

Solve.

3+4y=11

Subtract by 3 from both sides.

3+4y-3=11-3

Solve.

11-3=8

Rewrite the problem down.

4y=8

Divide by 4 from both sides.

4y/4=8/4

Solve.

8/4=2

y=2

x=1+3y

x=1+3*2

Solve.

PEMDAS stands for:

1+3*2

Multiply.

3*2=6

Add.

1+6=7

x=7

I hope this helps you! Let me know if my answer is wrong or not.

Step-by-step explanation:

Only 3 numbers are possible at units place (2, 4 & 6) as we need even number.

But at tens & hundreds place all 6 are possible

Number of 3 digit even numbers

= 3x6x6 = 108

36

[tex]Solution:[/tex]

[tex]6\times6=36[/tex]

[tex]such\ as:[/tex]

[tex]11\ \ \ 22\ \ \ 33\ \ \ 44\ \ \ 55\ \ \ 66\\ 12\ \ \ 21\ \ \ 31\ \ \ 41\ \ \ 51\ \ \ 61\\ 13\ \ \ 23\ \ \ 32\ \ \ 42\ \ \ 52\ \ \ 62\\ 14\ \ \ 24\ \ \ 34\ \ \ 43\ \ \ 53\ \ \ 63\\ 15\ \ \ 25\ \ \ 35\ \ \ 45\ \ \ 54\ \ \ 64\\ 16+263646+56\\ 6+6+6+6+6+6\\ =36.[/tex]

I hope this helps you

:)

As per linear equation, there are 372 more girls in school than boys.

"A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is a variable, A is a coefficient and B is constant."

Given, there are 1970 pupils.

There are 799 boys in the school.

Therefore, the number of girls in the school is

[tex]= (1970-799)\\= 1171[/tex]

Therefore, the number of girls more than boys in the school is

[tex]= (1171-799)\\= 372[/tex]

In that school , there are 372 more girls than total number of the boys.

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The equation to solve for the width of a rectangular frame that has a total area of 140 square inches is 4x²+32x-80=0.

The area of the rectangle is the product of its length and its breadth.

As it is given that the length of the rectangular frame is (2x+10), while the width of the rectangular frame is (2x+6). Also, it is mentioned that the area of the rectangular frame is 140 in².

Now we know the area is the product of length and breadth, therefore,

[tex]\rm \text{Area of rectangle} = length \times breadth[/tex]

[tex]140 = (2x+10)(2x+6)\\\\140 = 4x^2 + 12x + 20x + 60\\\\0=-140+4x^2 + 12x + 20x + 60\\\\4x^2 + 32 x -80=0[/tex]

Hence, the equation to solve for the width of a rectangular frame that has a total area of 140 square inches is 4x²+32x-80=0.

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Answer:

A.)  4x2 + 32x − 80 = 0

Step-by-step explanation:

Just took the test

The translation of points A(2, 3), B(–1, –4), and C(2, –2), 1 unit left and 4 units down gives A'(1, -1), B'(-2, -8) and C'(1, -6)

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, reflection and dilation.

Given the points A(2, 3), B(–1, –4), and C(2, –2), the translation 1 unit left and 4 units down gives:

A'(1, -1), B'(-2, -8) and C'(1, -6)

The translation of points A(2, 3), B(–1, –4), and C(2, –2), 1 unit left and 4 units down gives A'(1, -1), B'(-2, -8) and C'(1, -6)

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