someone help me please with this algebra problem

Answer:

894

Step-by-step explanation:

this can be "split" into 3 rectangles.

their areas can be easily calculated. and then we simply sum them all up for the answer.

rectangle 1 on the top

R1 = 5×9 = 45

rectangle 2 in the middle

R2 = (19+5)×(35-9-15) = 24×11 = 264

rectangle 3 at the bottom

R3 = 39×15 = 585

and all together

45+264+585 = 894

I think it's answer is 79 just guess

Answer:

slope = - [tex]\frac{6}{7}[/tex]

Step-by-step explanation:

Calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = ([tex]\frac{3}{2}[/tex], 2)

m = [tex]\frac{2-5}{\frac{3}{2}-(-2) }[/tex]

   = [tex]\frac{-3}{\frac{3}{2}+2 }[/tex]

   = [tex]\frac{-3}{\frac{7}{2} }[/tex]

  = - 3  × [tex]\frac{2}{7}[/tex]

  = - [tex]\frac{6}{7}[/tex]

Answer:

x = 5

Step-by-step explanation:

comment if you need explanation

Answer:

Step-by-step explanation:

It just so happens that x is the hypotenuse in the right triangle with sides 3 and 4. To find x we use Pythagorean's Theorem:

[tex]x^2=3^2+4^2[/tex] and

[tex]x^2=9+16[/tex] and

[tex]x^2=25[/tex] so

x = 5

Answer:

Original position: base is 1.5 meters away from the wall and the vertical distance from the top end to the ground let it be y and length of the ladder be L.

Step-by-step explanation:

By pythagorean theorem, L^2=y^2+(1.5)^2=y^2+2.25 Eq1.

Final position: base is 2 meters away, and the vertical distance from top end to the ground is y - 0.25 because it falls down the wall 0.25 meters and length of the ladder is also L.

By pythagorean theorem, L^2=(y -0.25)^2+(2)^2=y^2–0.5y+ 0.0625+4=y^2–0.5y+4.0625 Eq 2.

Equating both Eq 1 and Eq 2: y^2+2.25=y^2–0.5y+4.0625

y^2-y^2+0.5y+2.25–4.0625=0

0.5y- 1.8125=0

0.5y=1.8125

y=1.8125/0.5= 3.625

Using Eq 1: L^2=(3.625)^2+2.25=15.390625, L=(15.390625)^1/2= 3.92 meters length of ladder

Using Eq 2: L^2=(3.625)^2–0.5(3.625)+4.0625

L^2=13.140625–0.90625+4.0615=15.390625

L= (15.390625)^1/2= 3.92 meters length of ladder

hope it helps...

correct me if I'm wrong...

Answer:

I think it's 155 cm

Step-by-step explanation:

=(5×5×3)+(5×2×4)

= 75+40

= 155 cm2

Answer:

[tex]=2x^2+3xy-2y^2[/tex]

Step-by-step explanation:

When given the following problem;

[tex](2x-y)(x+2y)[/tex]

Distribute, multiply every number in one of the parenthesis by every number in the other;

[tex](2x-y)(x+2y)\\=(2x)(x)+(2x)(2y)+(-y)(x)+(2y)(-y)[/tex]

Simplify,

[tex]=(2x)(x)+(2x)(2y)+(-y)(x)+(2y)(-y)\\=2x^2+4xy-xy-2y^2\\=2x^2+3xy-2y^2[/tex]

Therefore, the final answer is;

[tex]=2x^2+3xy-2y^2[/tex]

Answer:

Radical (20)

Step-by-step explanation:

Radical ( (4-6)² + (2-6)²)) =radical ( 4+16) = radical (20)

Answer:

no..........................

Answer:

1×2+1=3

3×2+1=7

7×2+1=15

15x2+1=31

31×2+1=63

40÷2+4=24

24÷2+4÷16

16÷2+4=12

12÷2+4=10

10÷2+4=9

Answer:

90

Step-by-step explanation:

You are looking for the highest common factor for 270 and 360.  

Factor the 2 numbers

270 = 2 * 5 * 3  * 3 * 3

360 = 2 * 2 * 2 * 3 * 3 * 5

Each of the numbers has a 5

Each of the numbers has two threes

Each of the numbers has one 2

So the answer is 2 * 3*3 * 5

Answer:

[tex]\rm\displaystyle 0,\pm\pi [/tex]

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

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

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

[tex] \rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma ) [/tex]

factor out tanγ:

[tex]\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)[/tex]

divide both sides by tanαtanβ-1 and that yields:

[tex]\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}[/tex]

multiply both numerator and denominator by-1 which yields:

[tex]\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)[/tex]

recall angle sum indentity of tan:

[tex]\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta ) [/tex]

let α+β be t and transform:

[tex]\rm\displaystyle \tan( \gamma ) = - \tan( t) [/tex]

remember that tan(t)=tan(t±kπ) so

[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi ) [/tex]

therefore when k is 1 we obtain:

[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi ) [/tex]

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi ) [/tex]

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm \pi [/tex]

isolate -α-β to left hand side and change its sign:

[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }[/tex]

when is 0:

[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 ) [/tex]

likewise by Opposite Angle Identity we obtain:

[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 ) [/tex]

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm 0 [/tex]

isolate -α-β to left hand side and change its sign:

[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }[/tex]

and we're done!

Answer:

-π, 0, and π

Step-by-step explanation:

You can solve for tan y :

tan y (tan a + tan B - 1) = tan a + tan y

Assuming tan a + tan B ≠ 1, we obtain

[tex]tan/y/=-\frac{tan/a/+tan/B/}{1-tan/a/tan/B/} =-tan(a+B)[/tex]

which implies that

y = -a - B + kπ

for some integer k. Thus

a + B + y = kπ

With the stated limitations, we can only have k = 0, k = 1 or k = -1. All cases are possible: we get k = 0 for a = B = y = 0; we get k = 1 when a, B, y are the angles of an acute triangle; and k = - 1 by taking the negatives of the previous cases.

It remains to analyze the case when "tan "a" tan B = 1, which is the same as saying that tan B = cot a = tan(π/2 - a), so

[tex]B=\frac{\pi }{2} - a + k\pi[/tex]

but with the given limitation we must have k = 0, because 0 < π/2 - a < π.

On the other hand we also need "tan "a" + tan B = 0, so B = - a + kπ, but again

k = 0, so we obtain

[tex]\frac{\pi }{2} - a=-a[/tex]

a contradiction.

Given:

The inequality is:

[tex]-3x-4<2[/tex]

To find:

The values that are solutions to the given inequality.

Solution:

We have,

[tex]-3x-4<2[/tex]

Adding 4 on both sides, we get

[tex]-3x-4+4<2+4[/tex]

[tex]-3x<6[/tex]

Divide both sides by -3 and change the inequality sign because -3 is a negative value.

[tex]\dfrac{-3x}{-3}>\dfrac{6}{-3}[/tex]

[tex]x>-2[/tex]

Therefore, all the real values greater than -2 are the solutions to the given inequality.

Answer:

52°

Step-by-step explanation:

The angle 38° and m∠1 equal 90°, so 90-38=52.

Answer:

thats is an obtuse so if 2 is 38 then 1 should be 120 but u add those together u get 158 so it shold be 120 but if not then try looking explamles up

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

This is because the answer you picked means congruent to, however these two triangles are not the same size

Answer:

24. A)14.2

25. D)5.5

Step-by-step explanation:

I hope it help for you keep safe

Answer:

See explanation

Step-by-step explanation:

Function A is not clear; I will use the following in place of function A

Function A:

[tex]x \to\ 1 |\ 3 |\ 4 |\ 6[/tex]

[tex]y \to -1|\ 3|\ 5|\ 9[/tex]

Function B:

[tex]y = 2x + 4[/tex]

Required

Compare both functions

For linear functions, we often compare the slope and the y intercepts only.

Calculating the slope of function A, we have:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Where:

[tex](x_1,y_1) = (1,-1)[/tex]

[tex](x_2,y_2) = (3,3)[/tex]

So, we have:

[tex]m = \frac{3 - -1}{3 - 1}[/tex]

[tex]m = \frac{4}{2}[/tex]

[tex]m = 2[/tex]

To calculate the y intercept, we set [tex]x = 0[/tex], then solve for y

i.e.[tex](x,y) = (0,y)[/tex]

Using the slope formula, we have:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

Where:

[tex]m = 2[/tex]

[tex](x_1,y_1) = (0,y)[/tex]

[tex](x_2,y_2) = (3,3)[/tex]

So, we have:

[tex]2 = \frac{3 - y}{3 - 0}[/tex]

[tex]2 = \frac{3 - y}{3}[/tex]

Multiply by 3

[tex]6 = 3 - y[/tex]

Collect like terms

[tex]y = 3 - 6[/tex]

[tex]y = -3[/tex]

So, for function A:

[tex]m = 2[/tex] -- slope

[tex]y = -3[/tex] --- y intercept

For function B

[tex]y = 2x + 4[/tex]

A linear function is represented as:

[tex]y = mx + b[/tex]

By comparison

[tex]m = 2[/tex] --- slope

[tex]b = 4[/tex] --- y intercept

By comparing the results of both functions, we have the following conclusion:

Functions A and B have the same slope (i.e. 2)

Function B has a greater y intercept (i.e. 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.

To know more about identity:

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Hello!

14.8 = n - 0.3

n - 0.3 = 14.8

n = 14.8 + 0.3

n = 15.1 → value of n

Good luck! :)

Answer:

D: n = 15.1

Step-by-step explanation:

14.8 = n minus 0.3

14.8= n - 0.3

14.8    - is +
+
0.3
=
15.1
:D if u need help with more just ask!
I love this kind of math.

Here we are provided with a diagram of a triangle. We need to find out the length of the segment LM . As we can see that ,

∆ KNL ≈ MNL , [ By AAS ]

Therefore ,

⇒ KN = MN

⇒ 14x - 3 = 25

⇒ 14x = 25 + 3

⇒ 14x = 28

⇒ x = 2

Put this x = 2 in LM :-

⇒ LM = 9x + 5

⇒ LM = 9*2 + 5

⇒ LM = 18 + 5

⇒ LM = 23

Hence the required answer is 23 .

Answer:

5 ..... 144/16 = 25 sqrt(25) = 5

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Option B is correct, 4 and 3/10 hours  will Mrs. Nygaard need to work over the weekend to finish grading the projects.

A fraction represents a part of a whole.

To find the total amount of time Mrs. Nygaard has available during the week to grade projects, we need to add up the hours for each day:

1 and three-fourths + 1 and one-half + 1 and one-fifth + 2 + 1 and one-fourth = 7 and five-tenths hours

This means that Mrs. Nygaard has 7.5 hours during the week to grade projects.

If she needs 12 hours to grade all the projects, then she will need to work for an additional:

12 - 7.5 = 4.5 hours over the weekend to finish grading the projects.

Therefore, 4 and 3/10 hours  required to Mrs. Nygaard need to work over the weekend to finish grading the projects.

To learn more on Fractions click:

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Answers:

3 chocolate cones

1 strawberry cones

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

Explanation:

Let's isolate the variable c in the first equation

c+s = 4

c = 4-s

we subtract s from both sides to get c all by itself. This will then be plugged into the second equation. I'm using the substitution property.

1.75c + 1.3s = 6.55

1.75(4-s) + 1.3s = 6.55 ..... replace c with 4-s

1.75*4 + 1.75*(-s) + 1.3s = 6.55

7 - 1.75s + 1.3s = 6.55

-0.45s + 7 = 6.55

-0.45s = 6.55 - 7

-0.45s = -0.45

s = -0.45/(-0.45)

s = 1

So he bought 1 strawberry cone. Use this value of s to find c

c = 4-s

c = 4-1

c = 3

This says he also bought 3 chocolate cones.

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

As a check, we see that c+s = 3+1 = 4, showing that he bought 4 cones total.

Also,

1.75c + 1.3s = 1.75*3 + 1.3*1 = 6.55

indicating he spent $6.55 total for the four cones. This matches with what the instructions tell us, so the answer is confirmed.

Answer:

Step-by-step explanation:

[tex]\left \{ {{c+s=4} \atop {1.75c+1.3s=6.55}} \right.[/tex]

1.75c + 1.75s = 1.75 × 4 ........ (1)

1.75c + 1.3s = 6.55 ........ (2)

(1) - (2)

0.45s = 0.45 ⇒ s = 1

c + 1 = 4 ⇒ c = 3

[ 3 ] chocolate  [ 1 ] strawberry

Answer:

the answer is D

Step-by-step explanation:

Answer:

I think album is another name of CD.

Answer: its 20 I think

Answer:

x = 50

I hope this help the side note also help me a lot as well

Step-by-step explanation:

7 people = 280 liters

1 p = 40 liters

50 p = 40 x 50

50 p = 2000 liters

hope it helps.

Also, I think that Brainly is an awesome app, but there's an app which is doing great work for me in maths, named Gauthmath. I will suggest it. Video concepts and answers from real tutors.

Answer:

[tex]m\angle A= 30^{\circ}[/tex]

Step-by-step explanation:

In the question, we're given [tex]\angle A\cong \angle D[/tex]. Therefore, the measure of these two angles must be equal.

To find the value of [tex]x[/tex], set these two angles equal to each other:

[tex]4x-26=x+16[/tex]

Add 26 and subtract [tex]x[/tex] from both sides:

[tex]3x=42[/tex]

Divide both sides by 3:

[tex]x=\frac{42}{3}=14[/tex]

Since [tex]\angle A[/tex] was labelled as [tex]4x-26[/tex], substitute [tex]x=14[/tex] to find its measure:

[tex]\angle A=4(14)-26,\\\angle A=56-26,\\\angle A=\boxed{30^{\circ}}[/tex]

You can also substitute [tex]x=14[/tex] into the label of angle D as angle A is congruent to angle D for easier calculations ([tex]14+16=\boxed{30^{\circ}}[/tex]).