Graph y= -3x+4
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Answer:

x = 5

Step-by-step explanation:

[tex]6x - 20 = 2x \\ - 20 = 2x - 6x \\ - 20 = - 4x \\ \frac{ - 20}{ - 4} = x = 5[/tex]

Answer:2x-1

Step-by-step explanation:

Answer:

C) y+6=-2(x-4)

Step-by-step explanation:

Answer:

x=14

y=41

x+y= 55

Step-by-step explanation:

Since a and b are parallel you know that 123° and 3y are the same degree, so 123 divided by 3 because its 3×y= 41 so y=41

the line123° and 4x+1 are on has to equal 180 because its a straight line, 180-123= 57. 57- 1 because its not with the x= 56. 56÷4 because its 4×x/its with the x= 14. Hope thsi helps :)

Answer:

60 cents per hour

Step-by-step explanation:

final price minus starting price is 3 divide by 5 hours to get .60

Answer:

y - 4 = -5(x + 1)

Step-by-step explanation:

Slope = m =  -5

(-1, 4)

y - y1 = m(x - x1)

y - 4 = -5(x - [-1])

y - 4 = -5(x + 1)

Answer:la cantidad de invitados y la cantidad de mesas que no nos dio por favor agréguelo para que pueda ayudarlo :))

Step-by-step explanation:

Answer:

-Scalene

- Isosceles

-Perpendicular lines

Since only one box contains gold, then the gold must be in Box 2.

Let's analyze the statements in each box:

Box 1 says, "The gold is not in Box 2." If this statement is true, it means the gold cannot be in Box 2.

Box 2 says, "The gold is in this box." If this statement is true, it means the gold must be in Box 2.

Box 3 says, "The gold is not in this box." If this statement is true, it means the gold cannot be in Box 3.

Now, let's consider the possibilities:

If Box 1 is true, then the gold is not in Box 2 (as stated in Box 1), and it is also not in Box 3 (since Box 3 would be false, and only one box contains the gold).

As per Box's 1 statement, the gold is not in Box 2.

If Box 3 is true, then the gold is not in Box 3 (as stated in Box 3), and it is also not in Box 2 (since Box 2 would be false, and only one box contains the gold).

Therefore, the gold must be in Box 1.

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#SPJ3

The quantity of a 40% iodine solution that must be added to 40 gallons of a 15% iodine solution to make a 20% iodine solution is: 10 gallons.

According to the question;

Let gallons of a 40% iodine solution to be added be; x gallons.

Therefore, the total gallons of a 20% iodine solution obtained is; (40+x) gallons.

By mass balance of iodine in each solution;

In essence; 10 gallons of a 40% iodine solution must be added to make a 20% iodine solution.

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Answer:

1.5 Or C :)

Step-by-step explanation:

Answer:

12.5 cm^2

Step-by-step explanation:

5 × 2.5 = 12.5

...............

Answer:

AAS

Step-by-step explanation:

I'm pretty sure its this, but im sorry if im wrong!

Answer:

Adults (x) = 99 and

Seniors (y) = 393

393 seniors

Step-by-step explanation:

Adults= x

Seniors = y

1) x+ y = 492

2) 24x + 14y = 7,878

1) x = 492 - y

2) 24 (492 - y) + 14 y = 7,878 (substitution)

11808 - 24y + 14y = 7,878

11,808 - 10y = 7,878

11,808 - 7,878 = 10y

3,930/10 = y

y = 393

1) x + 393 = 492 (substitution)

x = 492 - 393

x = 99

Answer:

After turn one she will have 0 points !

Step-by-step explanation:

After turn one she will have 0 points

After turn 2 she will have 8 points

After turn 3 she will have 7 points

She will not earn bonus points because she could not reach ten points

(Hope this helps can I pls have brainlist (crown)☺️)

Answer:

1

Step-by-step explanation:

use BEDMAS here:

4 divided by 8 is 1/2 or 0.5

2 divided by 4 is 1/2 or 0.5

now adding these two gives you:

1/2 + 1/2 = 1

Answer:

60

Step-by-step explanation:

S₄ = 4(4+1)(4+2) = 120

S₃ = 3(3+1)(3+2) = 60

4th term = 120-60=60

Answer:

47%

Step-by-step explanation: 375 × .125 = .046875

rounding up, 5 keeps the 7 as is, 7 is greater than 5, therefore the 8, becomes a 9, 9 is greater than 5, so the 6 becomes a 7, .047.

move the decimal three digits to the left and you get, 47%.

Step-by-step explanation:

......

Answer:

f = 3

Step-by-step explanation:

7f + 6 = 27 ( subtract 6 from both sides )

7f = 21 ( divide both sides by 7 )

f = 3

7f+6=27

7f=27-6

7f=21

Divide both sides by 7 to isolate f:

f=3

Hope this helps!

#HaveAnAmazingDay

[tex]GraceRosalia[/tex]

Answer:

Sale kute bhan के lole

Tera baap aya ajju(^bhai 94

And???? What are we computing, you just gave me a fact.

___________

[tex] \: [/tex]

[tex] {4}^{4} ( {4}^{ - 7} )(4) = ...[/tex]

[tex] = {4}^{4 + ( - 7) + 1} [/tex]

[tex] = {4}^{ - 3 + 1} [/tex]

[tex] = \bold{{4}^{ - 2} } \to({ \sf{d.}})[/tex]

Hello Carousel!

[tex] \huge \boxed{\mathbb{QUESTION} \downarrow}[/tex]

[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]

[tex] \tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex]

First, let's take I as ⇨ [tex]\tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex].

[tex]\tt \:I = \int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x +1}} \: dx \\\tt \: I =\int \: \frac{ x + 2 }{( {x}^{2} + 2x + 1 + x + 2) \sqrt{x + 1}} \: dx \\ \tt \:I =\int \: \frac{ x + 2 }{( ({x + 1}^{2}) + x + 2) \sqrt{x + 1}}[/tex]

Let, x + 1 = m² => dx = 2mdm.

[tex]\tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: m} 2mdm \\ \tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: \bcancel{ m}} 2 \bcancel{m}dm \\ \tt \: I = \: \int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 } 2dm[/tex]

Now, divide the numerator & denominator by m²....we'll get it as...

[tex]\tt \:I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{ {m}^{2} + 1 + \frac{1}{ {m}^{2} } } \: dm \\\tt \: I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{( {m}^{2} + \frac{1}{ {m}^{2} } - 2) + 3} \: dm \\\tt \: I =2\int \: \frac{ (1 + \frac{1}{ {m}^{2} }) \: dm }{ ({m} - \frac{1}{ m } ) ^{2} + 3} [/tex]

Now, let m - 1/m be t => (1 + 1/m²) dm = dt

[tex]\tt \:I =2\int \: \frac{ dt}{ {t}^{2} + 3 } \\ \tt \:I = 2\int \: \frac{ dt}{ {t}^{2} + ( \sqrt{3}) ^{2} }[/tex]

We know, [tex]\tt \:\int \: \frac{dx}{ {x}^{2} + a ^{2} } = \frac{1}{a} tan ^{ - 1} (\frac{x}{a} ) + c[/tex]...therefore...

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{t}{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 1}[/tex]

Now, substitute the value of 't' in eq. 1..we'll get..

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{m - \frac{1}{m} }{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 2}[/tex]

Now, substitute the value of 'm' in eq. 2...we'll get...

[tex] \tt \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{ \sqrt{x + 1} - \frac{1}{ \sqrt{x - 1} } }{ \sqrt{3} } ) + c \\ \tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x + 1 - 1}{ \sqrt{3} \sqrt{x - 1} } ) + c \\ \boxed{\boxed{ \bf \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x }{ \sqrt{3 (x - 1} )} ) + c }}[/tex]

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Hope it'll help you!

ℓu¢αzz ッ

Answer:

k=4+h

...................

The probability distribution for all possible values of X are given in the discrete probability table ;

Fair coin = {H, T}

card = {J, Q, K, A}

Sample space :

Probability of winning ; P(Winning) = 2/8 = 1/4 = 0.25

Probability of not winning ;

1 - P(wininng) = 1 - 1/4 = 3/4 = 0.75

X : ____ winning ____ not winning

P(X) : ____ 0.25 ________ 0.75

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