Evaluate the price wise function ag the indicated value from the domain:

A. f(-3)=-1/2
B. f(-3)=-1/3
C.f(-3)=4
D.f(-3)=9

[tex]\displaystyle\bf (3-5\sqrt{5} )-(5\sqrt{3} -3)=3+3-5\sqrt{5} -5\sqrt{3} =\boxed{6-5\sqrt{5} -5\sqrt{3} } \\\\\\\\\sqrt{5} (\sqrt{5} -2x)=\boxed{-2x\sqrt{5} +5}[/tex]

Answer:

1134

Step-by-step explanation:

We have 3 digits:

a, b, c

a 3 digit number can be written as:

a*100 + b*10 + c*1

Such that these numbers can be:

{1, 2, 3, 4, 5, 6, 7, 8, 9}

Let's assume that:

a < b < c

Then the 3 smaller numbers are:

a*100 + b*10 + c

a*100 + c*10 + b

b*100 + a*10 + c

The 3 larger numbers are:

b*100 + c*10 + a

c*100 + a*10 + b

c*100 + b*10 + a

We know that the sum of the 3 smaller numbers is equal to 540, then:

(a*100 + b*10 + c) + (a*100 + c*10 + b) + (b*100 + a*10 + c) = 540

Let's simplify this:

(a + a + b)*100 + (b + c + a)*10 + (c + b + c) = 540

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

The sum of the 3 larger numbers is equal to X, we want to find the value of X:

(b*100 + c*10 + a) + (c*100 + a*10 + b) + (c*100 + b*10 + a) = X

Now let's simplify the left side:

(b + c + c)*100 + (c + a + b)*10 + (a + b + a)*1 = X

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Then we have two equations:

(2a + b)*100 + (b + c + a)*10 + (2c + b) = 540

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

Notice that the terms are inverted.

By looking at the first equation, we can see that:

(2c + b) = 10    (because the units digit of 540 is 0)

Then, we can see that:

(b + c + a + 1 ) = 14   (the one comes from the previous 10)

finally:

(2a + b + 1) = 5   (the one comes from the previous 14)

Then we can rewrite:

(2*c + b) = 10

(b + c + a) = 14 -1  = 13

(2a + b) = 5 - 1 = 4

Now we can replace these 3 in the equation:

(b + 2*c)*100 + (c + a + b)*10 + (2a + b) = X

(10)*100 + (13)*10 + 4 = X

1000 + 130 + 4 = X

1134 = X

The sum of the 3 largest numbers is 1134.

[tex]\color{red}{==========================}[/tex]

✍︎ꕥᴍᴀᴛʜᴅᴇᴍᴏɴǫᴜᴇᴇɴꕥ

✍︎ꕥᴄᴀʀʀʏᴏɴʟᴇᴀʀɴɪɴɢꕥ

Answer:

The answer is D (-0.7)

Step-by-step explanation:

Using d formular,a is 1,b is -5,c is -4

Input ur values and u will get ur answer

Answer:

OMG I LOVE LOUIS SO MUCH HE IS SO PERFECT

Step-by-step explanation:

Answer:

6.5

Step-by-step explanation:

We need to solve out for x , the given Equation is ,

[tex]: \implies[/tex] x - 3x + 2 = -11

[tex]: \implies[/tex] -2x = -11 -2

[tex]: \implies[/tex] -2x = -13

[tex]: \implies[/tex] x = -13/-2

[tex]: \implies[/tex] x = 6.5

Answer:

x = 6.5

Step-by-step explanation:

[tex]x - 3x + 2 = - 11 \\ - 2x + 2 = - 11 \\ - 2x = - 11 - 2 \\ \frac{ - 2x}{ - 2} = \frac{ - 13}{ - 2} \\ x = + 6.5 [/tex]

Answer: 7.7 miles

Step-by-step explanation:

2.5x + 5.75 = 25 (x = each mile traveled)

2.5x = 25 - 5.75

2.5x = 19.25

x = 7.7 miles

Answer:

$ 20d

Step-by-step explanation:

Since each Apple costs d dollars , therefore 20 will cost , $ 20d .

The correct expression that gives the cost of 20 apples, in dollars, is 20a/d. So, correct option is A.

To find the cost of 20 apples, we need to determine the expression that correctly calculates the cost based on the given variables.

Given that a represents the cost of one apple in dollars and d represents the number of dollars, we can identify the expression that calculates the cost of 20 apples.

Let's analyze each option:

A. 20a/d: This expression calculates the cost of 20 apples by multiplying the cost of one apple (a) by 20 and then dividing by d. Therefore, this expression correctly gives the cost of 20 apples in dollars.

B. 20d/a: This expression calculates the cost of 20 apples by multiplying the number of dollars (d) by 20 and then dividing by the cost of one apple (a). This does not give the correct cost of 20 apples.

C. a/20d: This expression calculates the cost of 20 apples by dividing the cost of one apple (a) by 20 and then dividing by d. This does not give the correct cost of 20 apples.

D. 20/ad: This expression calculates the cost of 20 apples by multiplying 20 by d and then dividing by the cost of one apple (a). This does not give the correct cost of 20 apples.

Therefore, the correct expression that gives the cost of 20 apples, in dollars, is A. 20a/d.

To learn more about equation/expression click on,

https://brainly.com/question/26372151

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Complete question is:

If a apples cost d dollars, which of the following expressions gives the cost of 20 apples, in dollars?

A. 20a/d

B. 20d/a

C. a/20d

D. 20/ad

First, we need to set up our two equations. For the picture of this scenario, there is one length (L) and two widths (W) because the beach removes one of the lengths. We will have a perimeter equation and an area equation.

P = L + 2W

A = L * W

Now that we have our equations, we need to plug in what we know, which is the 40m of rope.

40 = L + 2W

A = L * W

Then, we need to solve for one of the variables in the perimeter equation. I will solve for L.

L = 40 - 2W

Now, we can substitute the value for L into L in the area equation and get a quadratic equation.

A = W(40 - 2W)

A = -2W^2 - 40W

The maximum area will occur where the derivative equals 0, or at the absolute value of the x-value of the vertex of the parabola.

V = -b/2a

V = 40/2(2) = 40/4 = 10

Derivative:

-4w - 40 = 0

-4w = 40

w = |-10| = 10

To find the other dimension, use the perimeter equation.

40 = L + 2(10)

40 = L + 20

L = 20m

Therefore, the dimensions of the area are 10m by 20m.

Hope this helps!

Answer:

Width: 10 m

Length: 20 m

Step-by-step explanation:

Hi there!

Let w be equal to the width of the enclosure.

Let l be equal to the length of the enclosure.

1) Construct equations

[tex]A=lw[/tex] ⇒ A represents the area of the enclosure.

[tex]40=2w+l[/tex] ⇒ This represents the perimeter of the enclosure. Normally, P=2w+2l, but because one side isn't going to use any rope (sandy beach), we remove one side from this equation.

2) Isolate one of the variables in the second equation

[tex]40=2w+l[/tex]

Let's isolate l. Subtract 2w from both sides.

[tex]40-2w=2w+l-2w\\40-2w=l[/tex]

3) Plug the second equation into the first

[tex]A=lw\\A=(40-2w)w\\A=40w-2w^2\\A=-2w^2+40w[/tex]

Great! Now that we have a quadratic equation, we can do the following:

4) Solve for w

[tex]A=-2w^2+40w[/tex]

Factor out -2w

[tex]A=-2w(w-20)[/tex]

For A to equal 0, w=0 or w=20.

The average of 0 and 20 is 10, so the width that will max the area is 10 m.

5) Solve for l

[tex]40=2w+l[/tex]

Plug in 10 as w

[tex]40=2(10)+l\\40=20+l\\l=20[/tex]

Therefore, the length of 20 m will max the area.

I hope this helps!

Answer:

0.2 = n

Step-by-step explanation:

Since we want to express 3:15 as n:1, we can say that they are equal ratios, so

3:15 = n:1. Next, ratios can be written as fractions, so 3/15 = n/1 .

3/15 = n/1

Since anything divided by 1 equals itself, we can write this as

3/15 = n

0.2 = n

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:

(fg)(x) = (4x^2 + x^4)(x^2 + 4)^(1/2)

Step-by-step explanation:

Mathematically;

(fg)(x) = f(x) * g(x)

so we have;

4x^2 +x^4 * √(x^2 + 4)

But √(x^2 + 4) = (x^2 + 4)^(1/2)

So we have;

(fg)(x) = (4x^2 + x^4)(x^2 + 4)^(1/2)

Answer:

(0,1) is not the solution to the equation

Step-by-step explanation:

Given:

−3x − 8y = −8 (1)

y = 2 − x ​ (2)

Substitute (2) into (1)

−3x − 8y = −8 (1)

- 3x - 8(2 - x) = -8

- 3x - 16 + 8x = -8

- 3x + 8x = -8 + 16

5x = 8

x = 8/5

Substitute x = 8/5 into (2)

y = 2 − x ​ (2)

= 2 - 8/5

= (10-8) / 5

= 2/5

y = 2/5

Solution to the equation:

(x, y) = (8/5, 2/5)

Therefore,

(0,1) is not the solution to the equation

Answer:

hope it helps

Answer:

1. The third quartile

2. The quartile

3. Four

4. 25

5. The second quartile, Q₂

6. 20th percentile

7. Decile

8. Percentile

9. D₈

10. The third quartile, Q₃

Step-by-step explanation:

1. The other term(s) for Q₃ is third quartile (or upper quartile). It is the mid point between the distribution's largest number and the median

2. The measure of position divided into four equal parts is the quartile

3. The quartile is the division of the data at three points (Q₁, Q₂, and Q₃) into four equal parts equal parts

4. Q₁, which is the first quartile is equivalent to the 25th percentile

5. The other term for the median is the second quartile, Q₂

6. D₂ which is the second decile (a decile divides the data into 10 equal parts) is equivalent to 20th percentile

7. The decile, D, divides a given data into 10 equal parts

8. The percentile divides a given data into 100 equal parts

9. The decile equivalent to P80, which is the 80th percentile, is D₈

10. The quartile equivalent to P75, which is the 75th percentile or the three quarter mark point of the data is equivalent to the the third quartile, Q₃

There can be infinitely many pair of integers that have the sum -12.

Example:

1. (-24), (12)

2. (-10), (-2)

3. (-12), (0)

Answer:

Multiple pairs.

Step-by-step explanation:

There are a lot of solutions to this problem. For example:

(-24)+12

(-36)+24

(-48)+36

and so forth.

Answer:

2.56 kgs

Step-by-step explanation:

76.8kg ÷ 30 days = 2.56 kgs

Amount of sugar used - 76.8 kg

No. Of Days in Month of April - 30 days

∴ Sugar used each day is = 76.8 ÷ 30

[tex]76.8 \div 30 \\ \\ = \frac{76.8}{30} \\ \\ = \frac{768}{30 \times 10} \\ \\ = \frac{ \cancel{768}}{ \cancel{30} \times 10} \fbox{cancelling by 3} \\ \\ = \frac{256}{10 \times 10} \\ \\ = \frac{256}{100} \\ \\ = 2.56 \: kg[/tex]

Answer:

Option i

Step-by-step explanation:

The given expression is , 5x² - 3xy .

We can clearly see that there are two terms in the expression , 5x² and 3xy .

Therefore option (i) is correct .

Ax

B

Step-by-step explanation:

.....

...

..

Answer:

x = -4 ±3

x = -7,-1

Step-by-step explanation:

x^2 + 8x     = -7

1/2 B ^ 2 = 16

x^2 + 8x  + 16    = 16 -7

(x + 4) ^2 = 9

x + 4 = ±3

x = -4 ±3

x = -7,-1

Hello,

A:

[tex]\begin{array}{c|ccc|c}&x^3&x^2&x&1\\&1&-a&b&c\\x=-1&&-1&a+1&-a-b-1\\---&---&---&---&---\\&1&-a-1&a+b+1&-a-b+c-1\\\end{array}\\\\\\\begin{array}{c|ccc|c}&x^3&x^2&x&1\\&1&-a&b&c\\x=2&&2&-2a+4&-4a+2b+8\\---&---&---&---&---\\&1&-a+2&-2a+b+4&-4a+2b+c+8\\\end{array}\\\\\\\\-a-b+c-1=-4a+2b+c+8\\\\\boxed{b=a-3}\\[/tex]

B:

[tex](2x-1)^3+6(3+4x^2)\\\\=8x^3-12x^2+6x-1+18+24x^2\\\\=8x^3+12x^2+6x+17\\\\\\\begin{array}{c|ccc|c}&x^3&x^2&x&1\\&8&12&6&17\\x=-\dfrac{1}{2} &&-4&-4&-1\\---&---&---&---&---\\&8&8&2&16\\\end{array}\\\\\\(2x-1)^3+6(3+4x^2)=(2x+1)(4x^2+4x+1)+16[/tex]

Answer:

the answer is 200cm^3

explication

Answer:

Domain and range: [0, ∞)

Step-by-step explanation:

There is no limit to the number of concert tickets someone can buy, but there cannot be negative concert tickets. The domain, or what the input can be, is [0, ∞) as it can be greater than or equal to 0

The range is the total cost, or what the output can be. Since the total cost is the number of tickets multiplied by 55 (as it is $55 per ticket), this can be represented by 55 * number of tickets. The total cost cannot be negative, but it can be 0 if no one buys tickets. As there is no limit to the amount of tickets someone can buy, there is no limit to what the total cost could be, making the range [0, ∞)

Answer:

64

Step-by-step explanation:

2x2x2x2x2x2 = 64

Answer: Cant see the photo

Step-by-step explanation:

I cannot really see the photo

Answer:

[tex]{ \tt{f(x) = 2}} \\ when \: y = 2 \\ { \bf{x = - 0.5}}[/tex]