A school contains 140 boys and 160 girls. what is the ratio of boys to girls?
I need full working out please

Answer:

a. -3

Step-by-step explanation:

-2 turns into 6 by multiplying it by -3.

the same from 6 to -18, from -18 to 54, ...

so, an/an+1 = -3

Answer: 5 ft i think so

Use the Pythagorean theorem

Answer:

C'

Step-by-step explanation:

Given

ABCD to A'B'C'D'

Required

Corresponding angle of C

ABCD to A'B'C'D' means that the following angles are corresponding

[tex]A \to A'[/tex]

[tex]B \to B'[/tex]

[tex]C \to C'[/tex]

[tex]D \to D'[/tex]

Hence, C' corresponds to C

Answer:

C

Step-by-step explanation:

I took the test :)

Answer: [tex]7\sqrt{42}[/tex]

Step-by-step explanation:

42 students divided equally into 7 groups means 42 divided by 7, and [tex]7\sqrt{42}[/tex] is the only choice that shows that.

7√42. is the expressions would represent a class of 42 students divided equally into 7 groups

A division is a process of splitting a specific amount into equal parts.

Given,

A class of forty two Forty two students divided equally into 7 groups.

Forty two students divided equally into seven groups means forty two divided by seven, and

this can be done by using 7√42.

42 students divided equally into 7 groups means forty two divided by seven

Hence  7√42. is the expressions would represent a class of 42 students divided equally into 7 groups

To learn more on Division click:

https://brainly.com/question/21416852

#SPJ2

Answer:

A. (-2, -5), (0, -7), (1, -4)

Step-by-step explanation:

The following transformation is applied:

[tex](x,y) \rightarrow (y + 2, -x - 4)[/tex]

To find the range:

We apply the transformation to the points in the domain. Thus:

[tex](1,-4) \rightarrow (-4 + 2, -1 - 4) = (-2, -5)[/tex]

[tex](3,-2) \rightarrow (-2 + 2, -3 - 4) = (0, -7)[/tex]

[tex](0,-1) \rightarrow (-1 + 2, 0 - 4) = (1, -4)[/tex]

Thus the correct answer is given by option a.

Answer:

Buy 2 lollipops for $2.

Step-by-step explanation:

If you divide the total price by total items purchased, you get the price per unit. 2/2=1 or around $1, while 40/30=4/3 or around $1.3. You are paying 1$ per lollipop for the 2 lollipop choice and paying 1.3 dollars per lollipop for the 30 lollipop choice.

The requirement is fulfilled when 10 bananas and 2 apples are kept in the basket.

The equation can be given as

10y + 2x = 80

Here,

y means 7 ounces (weight of banasas)

x means 5 ounces (weight of apples)

Answer:

5 905 272

Step-by-step explanation:

you can refer to this lattice multiplication or u can search you tube for the examples of lattice multiplication

9514 1404 393

Answer:

Step-by-step explanation:

Of the items sold, sodas were 2/(2+1) = 2/3 of the total.

  (2/3)(228) = 152 . . . sodas were sold

  152/2 = 76 . . . . hot dogs were sold

Answer:

1. x + 4 = 9

Hint: the word 'sum' generally refers to addition.

2. 10a = 70

3. [tex]\frac{3}{4} t[/tex] = 15

4. [tex]\frac{1}{4} x[/tex] - 4 = 4

Answer:

12

Step-by-step explanation:

PEMDAS is the order

P = parenthesis

E = exponent

M = *

D = division

A = +

S = -

so first 8*2 = 16

and then -8/2 = -4

and then -4 + 16

= 12

Answer with Step-by-step explanation:

We are given that

Side of cube, x=30 cm

Error in measurement of edge,[tex]\delta x=0.5[/tex] cm

(a)

Volume of cube, [tex]V=x^3[/tex]

Using differential

[tex]dV=3x^2dx[/tex]

Substitute the values

[tex]dV=3(30)^2(0.5)[/tex]

[tex]dV=1350 cm^3[/tex]

Hence,  the maximum possible error in computing the volume of the cube

=[tex]1350 cm^3[/tex]

Volume of cube, [tex]V=(30)^3=27000 cm^3[/tex]

Relative error=[tex]\frac{dV}{V}=\frac{1350}{2700}[/tex]

Relative error=0.05

Percentage  error=[tex]0.05\times 100=5[/tex]%

Hence, relative error in computing the volume of the cube=0.05  and

percentage error in computing the volume of the cube=5%

(b)

Surface area of cube,[tex]A=6x^2[/tex]

[tex]dA=12xdx[/tex]

[tex]dA=12(30)(0.5)[/tex]

[tex]dA=180cm^2[/tex]

The maximum possible error in computing the volume of the cube=[tex]180cm^2[/tex]

[tex]A=6(30)^2=5400cm^2[/tex]

Relative error=[tex]\frac{dA}{A}=\frac{180}{5400}[/tex]

Relative error  in computing the volume of the cube=0.033

The percentage error in computing the volume of the cube=[tex]0.033\times 100=3.3[/tex]%

Answer:

4.17%

(1/4)(1/3)(1/2)(1)

alternative you can say that there are 24 permutations of

4 items and that you have to guess 1 of them 1/24 = 4.17%

Step-by-step explanation:

0.25  

0.333333333  

0.5  

1  

Answer:

1) has bigger answer

Step-by-step explanation:

1)

solving parenthesis first we get

5 × 10

so, the answer = 50

2)

solving 3 × 5 first as we have to see multiplication first then addition

2 + 15 + 5

22

comparing both

50 > 22

so problem 1 has a bigger answer

Answer:

B

Step-by-step explanation:

The coeffecients (I totally didn't spell that right) and variables match up.

Answer:

-2/3

Step-by-step explanation:

A rational number is a number that can be expressed by a fraction so when y add to fractions it’s a rational number.

Answer:

[tex]14x cos(\frac{1}{15}x^{2})=14 \sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]

Step-by-step explanation:

In order to find this Maclaurin series, we can start by using a known Maclaurin series and modify it according to our function. A pretty regular Maclaurin series is the cos series, where:

[tex]cos(x)=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{2k}}{(2k)!}[/tex]

So all we need to do is include the additional modifications to the series, for example, the angle of our current function is: [tex]\frac{1}{15}x^{2}[/tex] so for

[tex]cos(\frac{1}{15}x^{2})[/tex]

the modified series will look like this:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15}x^{2})^{2k}}{(2k)!}[/tex]

So we can use some algebra to simplify the series:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}(\frac{1}{15^{2k}}x^{4k})}{(2k)!}[/tex]

which can be rewritten like this:

[tex]cos(\frac{1}{15}x^{2})=\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]

So finally, we can multiply a 14x to the series so we get:

[tex]14xcos(\frac{1}{15}x^{2})=14x\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k}}{(2k)!15^{2k}}[/tex]

We can input the x into the series by using power rules so we get:

[tex]14xcos(\frac{1}{15}x^{2})=14\sum _{k=0} ^{\infty} \frac{(-1)^{k}x^{4k+1}}{(2k)!15^{2k}}[/tex]

And that will be our answer.

Answer:

34m = c

Step-by-step explanation:

For every month (m) you pay 34 dollars. However many months youu use that service time 34 equals your total cost (c).

Answer:

[tex]let \: cost \: be \: { \bf{c}} \: and \: months \: be \: { \bf{n}} \\ { \bf{c \: \alpha \: n}} \\ { \bf{c = kn}} \\ 34 = (k \times 1) \\ k = 34 \: dollars \\ \\ { \boxed{ \bf{c = 34n}}}[/tex]

Answer:

a) Figure B and Figure C

b) Figure C

c) Figure A, Figure B, and Figure C

Step-by-step explanation:

a) Rectangles are shapes that have four sides, and four right angels. Right angles are angles that are 90 degrees.

The only shapes with four sides AND four right angles are Figure C and Figure B.

b) Squares are shapes with four EQUAL sides and four right angles. The only shape with four equal sides and four right angles is Figure C.

c) Parallelograms are any shapes with four sides. All of these shapes have four sides.

Hope this helps!

Answer:

The angle for the other interior angel is 75°, all you have to do is subtract 180, from the 105

Step-by-step explanation:

Answer:

3x+4

Step-by-step explanation:

When you factor 9x^2+24x+16, it factors to (3x+4)^2

Factoring 9x^2 - 16 factors to (3x+4)(3x-4)

Therefore the common factor is 3x+4

I hope this helps!

Answer:

Area of ellipse=[tex]\pi ab[/tex]

Step-by-step explanation:

We are given that

[tex]x=acos\theta[/tex]

[tex]y=bsin\theta[/tex]

[tex]0\leq\theta\leq 2\pi[/tex]

We have to find the area enclose by it.

[tex]x/a=cos\theta, y/b=sin\theta[/tex]

[tex]sin^2\theta+cos^2\theta=x^2/a^2+y^2/b^2[/tex]

Using the formula

[tex]sin^2x+cos^2x=1[/tex]

[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

This is the equation of ellipse.

Area of ellipse

=[tex]4\int_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}dx[/tex]

When x=0,[tex]\theta=\pi/2[/tex]

When x=a, [tex]\theta=0[/tex]

Using the formula

Area of ellipse

=[tex]\frac{4b}{a}\int_{\pi/2}^{0}\sqrt{a^2-a^2cos^2\theta}(-asin\theta)d\theta[/tex]

Area of ellipse=[tex]-4ba\int_{\pi/2}^{0}\sqrt{1-cos^2\theta}(sin\theta)d\theta[/tex]

Area of ellipse=[tex]-4ba\int_{\pi/2}^{0} sin^2\theta d\theta[/tex]

Area of ellipse=[tex]-2ba\int_{\pi/2}^{0}(2sin^2\theta)d\theta[/tex]

Area of ellipse=[tex]-2ba\int_{\pi/2}^{0}(1-cos2\theta)d\theta[/tex]

Using the formula

[tex]1-cos2\theta=2sin^2\theta[/tex]

Area of ellipse=[tex]-2ba[\theta-1/2sin(2\theta)]^{0}_{\pi/2}[/tex]

Area of ellipse[tex]=-2ba(-\pi/2-0)[/tex]

Area of ellipse=[tex]\pi ab[/tex]

Answer:

7 brown M&Ms.

Step-by-step explanation:

This question is not complete, but I will assume that the final question is how many brown M&Ms will be in this package.

0.14 × 52 is our equation.

The answer is 7.28. We cannot have .28 of a brown M&M in a package (unless you count the broken ones) so there will be, on average, 7 brown M&Ms in a package.

(again, the question is incomplete, so this may not be the answer)

Answer:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Step-by-step explanation:

Given

[tex]f(x,y) = \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Required

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex]

[tex]\lim_{(x,y) \to (0,0)} f(x,y)[/tex] becomes

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}[/tex]

Multiply by 1

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2}\cdot 1[/tex]

Express 1 as

[tex]\frac{y^2}{y^2} = 1[/tex]

So, the expression becomes:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} \cdot \frac{y^2}{y^2}[/tex]

Rewrite as:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} \cdot \frac{\sin^2y}{y^2}[/tex]

In limits:

[tex]\lim_{(x,y) \to (0,0)} \frac{\sin^2y}{y^2} \to 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2} *1[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+2y^2}[/tex]

Convert to polar coordinates; such that:

[tex]x = r\cos\theta;\ \ y = r\sin\theta;[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{(r\cos\theta)^2 (r\sin\theta;)^2}{(r\cos\theta)^2+2(r\sin\theta;)^2}[/tex]

Expand

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2\cos^2\theta+2r^2\sin^2\theta}[/tex]

Factor out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^4\cos^2\theta\sin^2\theta}{r^2(\cos^2\theta+2\sin^2\theta)}[/tex]

Cancel out [tex]r^2[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+2\sin^2\theta}[/tex]

Express [tex]2\sin^2 \theta[/tex] as [tex]\sin^2\theta+\sin^2\theta[/tex]

So:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{\cos^2\theta+\sin^2\theta+\sin^2\theta}[/tex]

In trigonometry:

[tex]\cos^2\theta + \sin^2\theta = 1[/tex]

So, we have:

[tex]\lim_{(x,y) \to (0,0)} \frac{r^2\cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

Evaluate the limits by substituting 0 for r

[tex]\frac{0^2 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0 \cdot \cos^2\theta\sin^2\theta}{1+\sin^2\theta}[/tex]

[tex]\frac{0}{1+\sin^2\theta}[/tex]

Since the denominator is non-zero; Then, the expression becomes 0 i.e.

[tex]\frac{0}{1+\sin^2\theta} = 0[/tex]

So,

[tex]\lim_{(x,y) \to (0,0)} \frac{x^2 \sin^2y}{x^2+2y^2} = 0[/tex]

Given:

The graph of a line.

To find:

The equation for the given line.

Solution:

From the given graph, it is clear that the line passes through the points (0,-5) and (5,0). So, the equation of the line is:

[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

[tex]y-(-5)=\dfrac{0-(-5)}{5-0}(x-0)[/tex]

[tex]y+5=\dfrac{5}{5}(x)[/tex]

[tex]y+5=x[/tex]

Subtract 5 from both sides.

[tex]y+5-5=x-5[/tex]

[tex]y=x-5[/tex]

Therefore, the equation of the given line is [tex]y=x-5[/tex].

Answer:

Step-by-step explanation:

P(4,3), Q(4,1), S(-1,3), R(-1,1)

[tex]Distance =\sqrt{(x_{2}-x_{1})^{2}+(y_{2} -y_{1})^{2}}\\\\PQ= \sqrt{(4-4)^{2}+(1-3)^{2}}\\\\=\sqrt{(-2)^{2}}\\\\\=\sqrt{4}\\\\=2 \ units\\\\\\QS=\sqrt{(-1-4)^{2}+(3-1)^{2}}\\\\=\sqrt{(-5)^{2}+(2)^{2}}\\\\=\sqrt{25+4}\\\\=\sqrt{29}\ units\\\\\\SR =\sqrt{(-1-[-1])^{2}+(1-3)^{2}}\\\\=\sqrt{(-1+1)^{2}+(-2)^{2}}\\\\=\sqrt{0+4}\\\\= \sqrt{4}\\\\= 2 \\\\\\PR = \sqrt{(-1-4)^{2}+(1-3)^{2}}\\\\=\sqrt{(-5)^{2}+(-2)^{2}}\\\\=\sqrt{25+4}\\\\=\sqrt{29}\\\\[/tex]

PQRS is a rectangle

Area= length *breadth

       = 2 * √29

       = 2√29 sq.units

Answer:

The system is inconsistent

Step-by-step explanation:

Given the matrix system solutions in the attachment, we can see that the coefficient of the matrix rank has become zero. This shows that it has no solution.

Answer:

Step-by-step explanation:

Since, CD is an altitude, ∠CDB will be a right angle.

m∠CDB = m∠CDA = 90°

By applying triangle sum theorem in ΔABC,

m∠CAB + m∠CBA + m∠ACB = 180°

20° + m∠CBA + 90° = 180°

m∠CBA = 180° - 110°

             = 70°

Therefore, m∠CBD = 70°

By applying triangle sum theorem in ΔBCD,

m∠BCD + m∠CDB + m∠DBC = 180°

m∠BCD + 90° + 70° = 180°

m∠BCD + 160° = 180°

m∠BCD = 20°

m∠CAD = m∠A = 20°

m∠ACD = 90° - m∠BCD

             = 90° - 20°

m∠ACD = 70°