I need help with question .

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]

Step-by-step explanation:

hope helps you

have a good day

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

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.

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Answer:

Are there certain options?

Step-by-step explanation:

2^5 is in other words 2x2x2x2x2

the answer to this is 32

Hope I helped?

Answer:

32

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.

~

[tex]{\color{red}{\huge{\underbrace{\overbrace{\mathfrak{\:\:\:\:\:\:\:꧁"Answer"꧂\:\:\: }}}}}}[/tex]

Because ∆XZW and WYZ is the same height and

Because ∆XZW and WYZ is the same height and the are of ∆WXZ to the the WYZ

[tex]so \:( \frac{1}{2}xw \times h)( \frac{1}{2} w)=7:2[/tex]

Because xy=21

[tex]so \: WY=2 \frac{1}{2}(7 + 2x2) \\ = 21 \frac{1}{2}p \times 2 \\ =\small\color{blue}{{{\boxed{\tt\red{} \:\:\:\:\:\:\:\:\:\: WY=\frac{14}{3}\:\:\:\: }}}}[/tex]

The length of the line segment WY is [tex]\frac{14}{3}[/tex] unit.

The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle.

Area of triangle = (1/2)base × height

According to the given question.

The ratio of the area of triangle WXZ to the area of triangle WYZ is 7:2.

Since, the height WZ for both the triangles WXZ and WYZ is same.

Let, WZ = h

Therefore, the ratio of the area of the triangles is given by  

 [tex]\frac{\frac{1}{2}WX(h) }{\frac{1}{2}(WY)(h) } =\frac{7}{2}[/tex]

⇒ [tex]\frac{WX}{WY} =\frac{7}{2}[/tex]

⇒ [tex]WX = \frac{7}{2} WY..(i)[/tex]

Since, in the given figure

[tex]XY = WY + WX[/tex]

⇒ [tex]21 = WY + \frac{7}{2} WY[/tex]   (from i)

⇒ [tex]21 = \frac{9}{2}WY[/tex]

⇒ [tex]WY = \frac{14}{3}[/tex]

Hence, the length of the line segment WY is [tex]\frac{14}{3}[/tex] unit.

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Answer:

t =2 , 3

Step-by-step explanation:

s (t) = t^2 + 2 t + 5

p (t) = 11 + 3 t

(a) s (1) = 8

s (2) = 13

s (3) = 20

s (4) = 29

p (1) = 14

p (2) = 17

p (3) = 20

p (4) = 23

Now equate both of them

[tex]t^2 + 2t + 5 = 11 + 3 t \\\\t^2 - t - 6 =0 \\\\t^2 - 3 t + 2t - 6 =0\\\\t(t - 3) + 2 (t - 3) = 0\\\\(t -3)(t-2)=0\\\\t =3, 2[/tex]

(b) It shows that the values are same at = 2 and t = 3.

Domain of the function

Step-by-step explanation:

149......hshdbhdhsbhsjsusvshhs

Answer: $15

Step-by-step explanation:

-$50 + $75 = $15

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:

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°.

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:

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:

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

======================================================

Explanation:

The notation <-5,2> is the same as writing the translation rule [tex](x,y) \to (x-5,y+2)[/tex]

It says: move 5 units to the left and 2 units up

The point (6,2) moves to (1,2) when moving five units to the left. Then it ultimately arrives at (1, 4) after moving 2 units up. You could move 2 units up first and then 5 units to the left later on, and you'd still arrive at (1, 4). In this case, the order doesn't matter (some combinations of transformations this won't be the case and order will matter).

---------

Or you could write out the steps like so

[tex](x,y) \to (x-5, y+2)\\\\(6,2) \to (6-5, 2+2)\\\\(6,2) \to (1, 4)\\\\[/tex]

We see that (6,2) moves to (1, 4)

Answer:

Check off both of the boxes

Step-by-step explanation:

In your reponse you included that most ancient societies didn't have symbols for math operations and didn't have symbols for variables/unknown quantities.

Hope it helps (●'◡'●)

Answer: x = 100 duck

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

the answer is C, comment if you need explanation

Step-by-step explanation:

Answer:

4w-8

Step-by-step explanation:

4(w-2)

(4*w)-(4*2)

4w-8

Answer:

4w-8

Step-by-step explanation:

4(w - 2)

Distribute

4*w - 4*2

4w-8

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:

x is 66 degrees

Step-by-step explanation:

since its a isosceles, two of the angles should be the same.

Answer:

There is a non-removable discontinuity at x = 3

Step-by-step explanation:

We are  given that

[tex]f(x)=\frac{x^2+9}{x-3}[/tex]

We have to find true statement about given function.

[tex]\lim_{x\rightarrow 3}f(x)=\lim_{x\rightarrow 3}\frac{x^2+9}{x-3}[/tex]

=[tex]\infty[/tex]

It is not removable discontinuity.

x-3=0

x=3

The function f(x) is not define at x=3. Therefore, the function f(x) is continuous for all real numbers except x=3.

Therefore, x=3 is non- removable discontinuity of function f(x).

Hence, option B is correct.

Answer:

The first choice, Equation A and equation C.

Step-by-step explanation:

The lines A and C are intersecting in the point (0,8). That is the solution for those lines.

Answer:

Step-by-step explanation:

0.18-0.32 = .25-1.95

-0.14 = 1.70

obviously that equation above is not true,  I suspect that there were some "x" variables on some of those numbers?

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:

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:

12, 120,012

Step-by-step explanation: