Graph: Y - 3 = 1/2 (x+2)

Answer:

P = 157.5 units

Step-by-step explanation:

Hello!

9x + 4 + x = 54 <=>

<=> 9x + x + 4 = 54 <=>

<=> 10x + 4 = 54 <=>

<=> 10x = 54 - 4 <=>

<=> 10x = 50 <=>

<=> x = 50 : 10 <=>

<=> x = 5 => 9 × 5 + 4 + 5 = 54

Good luck! :)

Answer:

x = 5

Step-by-step explanation:

First, combine like terms. Like terms are terms with the same variables as well as same amount of said variables:

9x + x + 4 = 54

(9x + x) + 4 = 54

10x + 4 = 54

Next, isolate the variable, x. Note the equal sign, what you do to one side, you do tot he other. Do the opposite of PEMDAS.

PEMDAS is the order of operations, and stands for:

Parenthesis

Exponents (& Roots)

Multiplication

Division

Addition

Subtraction

-

First, subtract 4 from both sides of the equation:

10x + 4 (-4) = 54 (-4)

10x = 54 - 4

10x = 50

Next, divide 10 from both sides of the equation:

(10x)/10 = (50)/10

x = 50/10 = 5

x = 5 is your answer.

~

Answer:

7 nickels, 5 quarters, 3 dimes

Step-by-step explanation:

7 nickels= 35 cents

5 quarters= $1.25

3 dimes= 30 cents

35+ 1.25+ 30= $1.90

Hope this helps!

Plz mark Brainliest if u can :)

Answer:

Step-by-step explanation:

The period of a trig function tells you how much space is taken up by one "up-down-up" of the graph, which is a revolution. Because half of this graph takes place in a span of 2π, then the whole graph will span 4π.

Answer:

Since both terms are perfect squares, factor using the difference of squares formula,  

a ^2 − b ^2 = ( a + b ) ( a − b )

where  

a = y

and  

b = 6  

( y + 6 ) ( y − 6 )

Answer: The ratio that represents the cosine of ∠T is [tex]\frac{56}{65}[/tex]

Step-by-step explanation:

We are given:

UV = 56 units

VT = 33 units

UT = 65 units

∠V = 90°

Cosine of an angle is equal to the ratio of base and the hypotenuse of the triangle. ΔTUV is drawn in the image below.

[tex]\cos \theta=\frac{\text{base}}{\text{hypotenuse}}[/tex]

Base of the triangle is UV and the hypotenuse of the triangle is TU

Putting values in above equation, we get:

[tex]\cos \theta=\frac{UV}{TU}=\frac{56}{65}[/tex]

Hence, the ratio that represents the cosine of ∠T is [tex]\frac{56}{65}[/tex]

Step-by-step explanation:

Answer:

SA=BA+LA and SA=BA+ph

Step-by-step explanation:

I just looked it up

The correct formulas to find the surface area of a right prism are:

SA = BA + LA and SA = BA + ph.

Options (A), and (D) are the correct answer.

A prism is a three-dimensional object.

There are triangular prism and rectangular prism.

We have,

The correct formulas to find the surface area of a right prism are:

1)

SA = BA + LA, where SA is the total surface area, BA is the area of the two identical bases, and LA is the lateral area (the sum of the areas of all the rectangular sides).

2)

SA = BA + ph, where SA is the total surface area, B is the area of one base, p is the perimeter of the base, h is the height of the prism, and ph is the area of all the rectangular sides (the lateral area).

Therefore,

The correct formulas to find the surface area of a right prism are:

SA = BA + LA and SA = BA + ph.

Learn more about prism here:

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Answer:

A.............

Step-by-step explanation:

. ..........

Answer:

C. (3,3)

Step-by-step explanation:

When These equations are both graphed the solution for these equations when they intersect is (-3,3)

Answer:

3/4

Step-by-step explanation:

tangent is opp/ad

so that is 15/20, or 3/4

Answer: (a), (b), (c), and (d)

Step-by-step explanation:

Check the options

[tex](a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x[/tex]

[tex](b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)[/tex]

[tex](c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x[/tex]

[tex](d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x[/tex]

Thus, all the identities are correct.

A. Not an identity

B. An identity

C. Not an identity

D. An identity

To check whether each expression is an identity, we need to verify if the equation holds true for all values of the variable x. If it is true for all values of x, then it is an identity. Let's check each option:

A. [tex]\((\sin x + \cos x)^2 = 1 + \sin 2x\)[/tex]

To check if this is an identity, let's expand the left-hand side (LHS):

[tex]\((\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x\)[/tex]

Now, we can use the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex] to simplify the LHS:

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

The simplified LHS is not equal to the right-hand side (RHS) 1 + sin 2x since it is missing the sin 2x term. Therefore, option A is not an identity.

B. [tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]

To check if this is an identity, we can use the double-angle identity for sine:[tex]\(\sin 2\theta = 2\sin \theta \cos \theta\)[/tex]

Let [tex]\(2\theta = 6x\)[/tex], which means [tex]\(\theta = 3x\):[/tex]

[tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]

The equation holds true with the double-angle identity, so option B is an identity.

C. [tex]\(\frac{\sin 3x}{\sin x \cos x} = 4\cos x - \sec x\)[/tex]

To check if this is an identity, we can simplify the right-hand side (RHS) using trigonometric identities.

Recall that [tex]\(\sec x = \frac{1}{\cos x}\):[/tex]

[tex]\(4\cos x - \sec x = 4\cos x - \frac{1}{\cos x} = \frac{4\cos^2 x - 1}{\cos x}\)[/tex]

Now, using the double-angle identity for sine, [tex]\(\sin 2\theta = 2\sin \theta \cos \theta\),[/tex] let [tex]\(\theta = x\):[/tex]

[tex]\(\sin 2x = 2\sin x \cos x\)[/tex]

Multiply both sides by 2: [tex]\(2\sin x \cos x = \sin 2x\)[/tex]

Now, the left-hand side (LHS) becomes:

[tex]\(\frac{\sin 3x}{\sin x \cos x} = \frac{\sin 2x}{\sin x \cos x}\)[/tex]

Using the double-angle identity for sine again, let [tex]\(2\theta = 2x\):[/tex]

[tex]\(\frac{\sin 2x}{\sin x \cos x} = \frac{2\sin x \cos x}{\sin x \cos x} = 2\)[/tex]

So, the LHS is 2, which is not equal to the RHS [tex]\(\frac{4\cos^2 x - 1}{\cos x}\)[/tex]. Therefore, option C is not an identity.

D. [tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x\)[/tex]

To check if this is an identity, we can use the sum-to-product trigonometric identities:

[tex]\(\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}\)\(\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\)[/tex]

Let A = 3x and B = x:

[tex]\(\sin 3x - \sin x = 2\sin x \cos 2x\)\(\cos 3x + \cos x = 2\cos 2x \cos x\)[/tex]

Now, we can rewrite the expression:

[tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \frac{2\sin x \cos 2x}{2\cos 2x \cos x} = \frac{\sin x}{\cos x} = \tan x\)[/tex]

The equation holds true, so option D is an identity.

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Answer: $15

Step-by-step explanation:

-$50 + $75 = $15

Answer:

(5, 4 )

Step-by-step explanation:

Given the 2 equations

3x - y = 11 → (1)

- 2x - 4y = - 26 → (2)

Multiplying (1) by - 4 and adding to (2) will eliminate the y- term

- 12x + 4y = - 44 → (3)

Add (2) and (3) term by term to eliminate y

- 14x + 0 = - 70

- 14x = - 70 ( divide both sides by - 14 )

x = 5

Substitute x = 5 into either of the 2 equations and solve for y

Substituting into (1)

3(5) - y = 11

15 - y = 11 ( subtract 15 from both sides )

- y = - 4 ( multiply both sides by - 1 )

y = 4

solution is (5, 4 )

Answer:

The answer is C.) m∠RCD = m∠ACD ÷ 2

RCD = ACD divided by two.

RCD = 90 degrees

ACD ÷2 = 180÷2 = 90 degrees.

So, your answer is C.

Hope this helped. Have a grey day!

Answer:

C. m∠RCD = m∠ACD ÷ 2

Hope this helps!

Step-by-step explanation:

I got it right.

Answer:

For perimeter you add up the side lengths to get the perimeter but for area you multiply the length times width (L x W )to get area.

Step-by-step explanation:

Answer:

Answer:

Radius of the circular garden

= 210 sq

=105m

Radius of the region covering the garden and the path =105m+7m

=112m

Area of the region between two concentric circles

with radius of outer circle R, and inner circle r =π(R sq−r sq)

Hence, the area of the path

=π(112sq−105 sq)= 7/22

(12544−11025)

= 33418/7

=4774m sq

HOPE THIS WILL HELP YOU MATE

Answer:

There are 4 terms

Step-by-step explanation:

A term is a single mathematical expression. Terms can be identified with a plus or minus sign in front of them. Terms can also be multiplied and divided.

So, in this case, the terms are:

-7

12x^4

-5y^8

x

Step-by-step explanation:

149......hshdbhdhsbhsjsusvshhs

Answer:

4th degree polynomial with 4 terms

Step-by-step explanation:

Given:

3n²(n²+ 4n - 5) - (2n² - n⁴ + 3)

Open parenthesis

= 3n⁴ + 12n³ - 15n² - 2n² + n⁴ - 3

Collect like terms

= 3n⁴ + n⁴ + 12n³ - 15n² - 2n² - 3

= 4n⁴ + 12n³ - 17n² - 3

Number 1 term is 4n²

Number 2 term is 12n³

Number 3 term is -17n³

Number 4 term is -3

The highest degree of the polynomial is 4th degree

Therefore,

The difference in 3n²(n²+ 4n - 5) - (2n² - n⁴ + 3) is

4th degree polynomial with 4 terms

Answer:

4th degree polynomial with 4 terms

Step-by-step explanation:

Answer:

Angle a = 80°, Angle b = 55°, Angle c = 45°, Angle d = 80°

Step-by-step explanation:

To find the measure of Angle a, we add 55 and 45, then subtract the sum from 180.

180 - 100 = 80

Angle a is 80°.

Then, we solve for Angle b. Line segment CD is congruent to Line AB, so Angle b is congruent to 55°.

After that, we find Angle c. Line segment AC is congruent to Line segment BD, so Angle c is congruent to 45°.

Lastly, we solve for Angle d using the same method we used for Angle b and Angle c. Angle d is congruent to Angle a, so it measures 80°.

So, Angle a = 80°, Angle b = 55°, Angle c = 45°, Angle d = 80°.

1cm = 2m

=> 1cm = 200cm

2.5cm = 2.5 × 200cm = 500 cm = 5m

So, the length of window is 500cm or 5m.

Answer:

First answer choice: [tex]y=\sqrt{x}[/tex] shifted right 6 units and down 2 units.

Step-by-step explanation:

Graph

Answer:

using Pythagoras' theorem c²=a²+b²

the diagonal is the hypotenuse of one of the triangles formed

let x represent one side of the square

√48²=x²+x²

√48²=2x²

48=2x²

48/2=2x²/2

24=x²

√24=√x²

4.8989794855663561=x

~4.90

Area of the square=side x side

4.90x4.90

24.01units²

Answer of this question

Answer:

U={a,b,c,d,e,f,g,h}

A={a,b,c,d}

now,

AUB= {a,b,c,d,e,f,g,h} U {a,b,c,d}

= {a,b,c,d,e,

ok,this question is wrong btw ,I think I stead of U(universal set) it is B

Answer:

First exercise

a) 1/8=0.125

b) 5$

c) y = 0.125 * x - 5

Second exercise

a) 0.11 $/KWh

b) no flat fee

c) y = 0.11 * x

Step-by-step explanation:

(See the pictures)

Answer:

12, 120,012

Step-by-step explanation:

Answer:

p = T - a - b

Step-by-step explanation:

T = a + p + b

p = T - a - b

Answer:

[tex]JL=54[/tex]

Step-by-step explanation:

We are given that K is the midpoint of JL. Using this information, we want to find JL.

By the definition of midpoint, this means that:

[tex]JK=KL[/tex]

Substitute them for their equations:

[tex]8x+11=14x-1[/tex]

Solve for x. Subtract 8x from both sides:

[tex]11=6x-1[/tex]

Add 1 to both sides:

[tex]6x=12[/tex]

And divide both sides by 6. Hence:

[tex]x=2[/tex]

JL is the sum of JK and KL. Hence:

[tex]JK+KL=JL[/tex]

Since JK = KL, substitute either one for the other:

[tex]JK+(JK)=2JK=JL[/tex]

Substitute JK for its equation:

[tex]2(8x+11)=JL[/tex]

Since we know that x = 2:

[tex]2(8(2)+11)=2(16+11)=2(27)=54=JL[/tex]

Thus:

[tex]JL=54[/tex]