The graph of g(x) resembles the graph of f(x) = x2, but it has been changed.
Which of these is the equation of g(x)?

Answer: [tex]30^{\circ}[/tex]

Step-by-step explanation:

[tex]\cos x=\frac{13}{15}\\\\x=\cos^{-1} \left(\frac{13}{15} \right)\\\\x \approx 30^{\circ}[/tex]

Step-by-step explanation:

you need to find:

uw:

uv:

vw= 8(already known)

Measure of W= 35

Measure of u= 90

measure of v= 180-90+35= 55

finding uv:

sin(35)= uv/8

0.57=uv/8

4.56=uv

finding uw:

sin(55)= uw/8

0.82= uw/8

6.56= uw

I think

x= -4, -1 and y=13, 8

(-4, 13) and (-1, 8)

Answer:

A triangular prism and a trapezoidal prism, if I understand the question

Step-by-step explanation:

There wound be a trapezoid and a triangle if you cut it horizontally. If you mean vertically, it would be a right triangular prism

220 cubic units.

Answer:

Solution given:

for small cylinder

r=1

and for large cylinder

R=5+1=6

height for both [h]=2

Now

Volume of solid=πR²h-πr²h=πh(R²-r²)

=3.14*2(6²-1²)=219.8 =220 units ³.

Small cylinder is r=1

Large cylinder is R= 5+1 =6

Height (h) =2

Volume of solid,

→ πR²h-πr²h

→ πh(R²-r²)

→ 3.14 × 2(6²-1²)

→ 219.8

→ 220 cubic units

Answer:

Step-by-step explanation:

1)

[tex]cot x sec^4 x = cotx + 2tanx + tan^3 x[/tex]

[tex]RHS =[/tex]

           [tex]=\frac{cosx}{sinx} + 2 \frac{sinx }{cosx} + \frac{sin^3x }{cos^3x}\\\\[/tex]

          [tex]= \frac{cosx(cos^3x) + 2sinx(sinx \ cos^2x ) + sin^3x(sinx)}{sinx \ cos^3x}\\\\[/tex]        [tex][\ taking LCM \ ][/tex]

         [tex]=\frac{cos^4x + 2sin^2xcos^2x +sin^4x }{sinx cos^3x}[/tex]

         [tex]= \frac{(sin^2 x + cos^2x)^2 }{sin x \ cos^3 x}[/tex]                        [tex][ \ a^4 + 2a^2b^2 + b^4 = (a^2 + b^2 ) ^2 \ ][/tex]

        [tex]= \frac{1}{sinx \ cos^3 x}\\\\= \frac{1 \times cosx}{sinx \times cos^3x \times cosx}[/tex]                    [tex][ \ multiplying\ and \ dividing \ by \ cosx \ ][/tex]

        [tex]= \frac{cosx}{sinx} \times \frac{1}{cos^4x}\\\\=cot x \ sec^4 x[/tex]

        [tex]= LHS[/tex]

2)

[tex]sin x \ ( tanx \ cosx - cotx \ cosx) = 1 - 2cos^2x[/tex]

[tex]LHS =[/tex]

        [tex]=sinx( [ \frac{sinx}{cosx} \times cosx)] - [ \frac{cosx}{sinx}\times cosx] )\\\\= sinx (sinx - \frac{cos^2x}{sinx})\\\\=sin^2x - cos^2 x\\\\=(1 - cos^2x ) -cos^2 x[/tex] [tex][ \ sin^2x = 1 -cos^2 x \ ][/tex]

        [tex]= 1 -cos^2x - cos^2 x \\\\= 1 - 2cos^2x \\\\=RHS[/tex]

3)

[tex]1 + sec^2x \ sin^2x = sec^2 x[/tex]

[tex]LHS =[/tex]

       [tex]= 1 +sec^2x \sin^2x \\\\= 1 + (\frac{1}{cos^2x} \times sin^2x )\\\\= 1 + \frac{sin^2 x}{cos^2x}\\\\= 1 + tan^2x \\\\= sec^2 x\\\\=RHS[/tex]

4)

[tex]\frac{sinx}{1 -cosx} + \frac{sinx}{1+cosx} = 2 \ cosec x[/tex]

[tex]LHS =[/tex]

[tex]=\frac{sinx}{1 -cosx} + \frac{sinx}{1+cosx} \\\\= \frac{sinx(1 +cosx)}{(1-cosx)(1+cosx)} + \frac{sinx(1-cox)}{(1+cosx)(1-cosx)}\\\\= \frac{sinx +sinx\ cosx}{(1 - cos^2x)} + \frac{sinx - sinx \ cosx}{1 - cos^2x}\\\\=\frac{sinx + sinx \ cosx + sinx - sinx \ cosx}{1 - cos^2x}\\\\=\frac{2sinx}{sin^2x}\\\\=\frac{2}{sinx}\\\\=2 cosec\ x\\\\=RHS[/tex]

5)

[tex]- tan^2 + sec^2 x = 1\\\\[/tex]

        [tex]sin^2 x + cos^2 x = 1\\\\\frac{sin^2x }{cos^2x} + \frac{cos^2x}{cos^x} = \frac{1}{cos^2x}\\\\tan^2x + 1 = sec^2x \\\\1 = sec^2 - tan^2x \\\\-tan^2x + sec^2 x = 1[/tex]                

Step-by-step explanation:

The given equation is :

y = 2x-7

We will first find the points,

x  0 , , 3.5, 1 , 2 , 3

y -7 , 0, -5, -3, -1

The graph of the above equation is as follows :

It is a linear equation.

Answer:

12

Step-by-step explanation:

bh/2

base is 8

height is 3

24/2

12

Answer:

12

Step-by-step explanation:

The formula to find the area of a triangle is the following,

[tex]A=\frac{b*h}{2}[/tex]

Where (b) and (h) are the base and height of the triangle. In this situation, one can see that the base (CB) equals (8) units. One can attain this number by counting the boxes between point (C) and point (B). The height is (3). This number is the distance from the point (A) to the side (CB). Substitute these values into the formula and solve for the area.

[tex]A=\frac{b*h}{2}[/tex]

Substitute,

[tex]A=\frac{8*3}{2}[/tex]

Simplify,

[tex]A=\frac{8*3}{2}\\\\A=\frac{24}{2}\\\\A=12[/tex]

I believe its <ABF =<EDJ

Answer:

Step-by-step explanation:

4). Given sequence → 3, 5, 7, 9, 11........

    Common difference of an arithmetic sequence is defined by,

    Common difference = Successive term - previous term

                                       = 2nd term - 1st term

                                       = 5 - 3

                                       = 2

5). Given sequence → 16, 19, 22, 25........

    Common difference of an arithmetic sequence = Second term - first term

                                                                                    = 19 - 16

                                                                                     = 3

Answer:

1x= 28

y= v7vub8b9nf7v6v

Answer:

0.088 is the answer...

Answer:

-63

Step-by-step explanation:

9 - (2×3) × 12

9 - (6 × 12)

9 - 72

= -63

Answer:

-63

Step-by-step explanation:

r − ns × 12

Let n = 2, r = 9, and s = 3

9 - (2)*3 * 12

Multiply first

9 - 72

-63

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is [tex]\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}[/tex].

Now we will solve this expression with the help of law of exponents.

[tex]\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}[/tex]

           [tex]=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}[/tex]

           [tex]=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}[/tex]

           [tex]=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}[/tex]

           [tex]=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}[/tex]

           [tex]=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}[/tex]

           [tex]=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }[/tex] [Option 2]

[tex]2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2[/tex] [Option 1]

[tex]2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2[/tex]

                [tex]=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }[/tex]

                [tex]=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }[/tex] [Option 3]

[tex]2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }[/tex]

               [tex]=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}[/tex] [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

Answer:x=( y-p-mq)/m

Step-by-step explanation:

y-p=m(x-q)

x-q= y-p/m

x=y-p/m=q= y-p+mq/m

x= y-p+mq/m

Answer:

You should first make sure you explain what exactly is the question, before delete others answers.‍♀️‍♀️

Step-by-step explanation:

Answer:

I think one tailed correct me if im wrong

Answer:

36/6 and 6/1 is equivalent

20/5 and 3/1 is not equivalent

18/3 and 9/1 is not equivalent

35/7 and 5/1 is equivalent

32/4 and 8/1 is equivalent

100/10 and 5/1 is not equivalent.

Step-by-step explanation:

36 divided by 6 is 6.

20 divided by 5 is 4.

18 divided by 3 is 6.

35 divers by 7 is 5.

32 divided by 4 is 8.

100 divided by 10 is 10.

Answer:

For every 100-kilogram increase in mass, the fuel efficiency of car decreases by about 1.1 miles per gallon.

Step-by-step explanation:

Fuel efficiency after x kg in mass:

y = 43.28 -0.011x

Thus, the slope is of -0.011 per x kg.

Increase of 100:

When x = 100, as the slope is negative, the efficiency will decay by:

-0.011*100 = -1.1

Decay of 1.1 miles per gallon, so the answer is:

For every 100-kilogram increase in mass, the fuel efficiency of car decreases by about 1.1 miles per gallon.

Answer:

it needs more information

Answer:

x = 9.7

Step-by-step explanation:

tan θ = opposite side / adjacent side

tan 37° = x / 13

multiply 13 on each side

13 × tan 37° = 13 × x /13

13 × tan 37° = x

Rewrite

x = 13 × tan 37°

multiply, we get

x = 13 × 0.75355..

x = 9.79620265

Round nearest tenth = 9.7

Answer:

The correct answer is "2.660".

Step-by-step explanation:

Given that,

Sample size,

n = 61

t-distribution with,

t = 0.005

Now,

The degree of freedom will be:

= [tex]n-1[/tex]

= [tex]61-1[/tex]

= [tex]60[/tex]

hence,

⇒ [tex]t,df=t_{0.005}, 60[/tex]

           [tex]=2.660[/tex]

Answer:

a - Right angled triangle

b - Parallelogram

c - Square

d - kite

Step-by-step explanation:

Hope this helps

Goodluck

These are the names of each of these shapes;

1. Right-angled triangle

2. Parallelogram

3. Square

4. kite

A right-angled triangle is a triangle, that has one of its interior angles equal to 90 degrees or anyone angle is a right angle.

A parallelogram is a quadrilateral with two pairs of parallel sides.

The opposite sides of a parallelogram are equal in length and the opposite angles are equal in measure.

A square is a four-sided polygon that has all sides equal in length and the measure of the angles is 90 degrees.

A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.

These are the names of each of these shapes;

1. Right-angled triangle

2. Parallelogram

3. Square

4. kite

Learn more about parallelogram here;

https://brainly.com/question/12906753

#SPJ2

Answer:

a) n and o

Step-by-step explanation:

by the concept of corresponding angles, angles that are equal when a line intersects 2 parallel lines .

l is the line that intersects n and o, and since x = 35, theyre corresponding angles resulting in n and o being parallel

Answer:

a) n and o

Step-by-step explanation:

Corresponding angles

Hope my answer helps uh ✌️

Answer:

The length of segment ZY is of [tex]\sqrt{73}[/tex] units.

Step-by-step explanation:

Distance between two points:

Suppose that we have two points, [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]. The distance between them is given by:

[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

In this question:

Point Z has coordinates (-5,1)

Pount Y has coordinates (3,4).

The length of segment ZY is the distance between points Z and Y. Thus

[tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]D = \sqrt{(-5-3)^2+(1-4)^2}[/tex]

[tex]D = \sqrt{8^2+3^2}[/tex]

[tex]D = \sqrt{73}[/tex]

The length of segment ZY is of [tex]\sqrt{73}[/tex] units.