Someone please help me with this math problem? !!

Answer:

832.5

A unit rate is a rate with 1 in the denominator.

The coordinates that describe the point D are: Option C: (0, 4)

The graph shows that the vertical axis is labelled as the y-axis while the horizontal axis is labelled as the x -axis.

Normally, the coordinate system of writing the given points of the graph is (x, y)

Thus, the point D on the graph is seem to be 4 units on the positive y-axis and 0 units on the positive x-axis and as such, we can say that the coordinates of point D are:

D(0, 4)

Read more about Graph Coordinates at: https://brainly.com/question/11337174

Complete question is:

Which of the following describes point D?

(-4, 0)

(0, -4)

(0, 4)

(4, 0)

Given:

The number of ribbons it can sell each week, x, is related to the price p per ribbon by the equation:

[tex]x=1000-100p[/tex]

To find:

The selling price if the company wants the weekly revenue to be $1,600.

Solution:

We know that the revenue is the product of quantity and price.

[tex]R=xp[/tex]

[tex]R=(1000-100p)p[/tex]

[tex]R=1000p-100p^2[/tex]

We need to find the value of p when the value of R is $1600.

[tex]1600=1000p-100p^2[/tex]

[tex]1600-1000p+100p^2=0[/tex]

[tex]100(16-10p+p^2)=0[/tex]

Divide both sides by 100.

[tex]p^2-10p+16=0[/tex]

Splitting the middle term, we get

[tex]p^2-8p-2p+16=0[/tex]

[tex]p(p-8)-2(p-8)=0[/tex]

[tex](p-8)(p-2)=0[/tex]

Using zero product property, we get

[tex]p-8=0[/tex] or [tex]p-2=0[/tex]

[tex]p=8[/tex] or [tex]p=2[/tex]

Therefore, the smaller value of p is $2 and the larger value of p is $8.

Hi there!

[tex]\large\boxed{73m^2}}[/tex]

Once again, divide the figure into 3 rectangles:

Top rectangle:

3m × 5m = 15m²

Long rectangle (subtract 5m from 9m to get the width):

12m × 4m = 48m²

Bottom rectangle:

2m × 5m = 10m²

Add up areas:

15m² + 48m² + 10m² = 73m²

Answer:

C

Step-by-step explanation:

The nth term of an AP is

[tex]a_{n}[/tex] = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = - 5 and d = a₂ - a₁ = - 10 - (- 5) = - 10 + 5 = - 5 , then

a₁₀ = - 5 + (9 × - 5) = - 5 - 45 = - 50 → C

[tex]\huge\boxed{\boxed{\underline{\textsf{\textbf{Answer}}}}}[/tex]

Number of days Adya took to complete the work = 20

Work done by Adya in 1 day = [tex]\frac{1}{20}[/tex]

Number of days Ashley took to complete the work = 25

Work done by Ashley in 1 day = [tex]\frac{1}{25}[/tex]

So,

Total work by Adya & Ashley in 1 day =

[tex] \frac{1}{20} + \frac{1}{20} \\ = \frac{5 + 4}{100} \\ = \frac{9}{100} [/tex]

•°• Their total work in 10 days =

[tex] \frac{9 \times 10}{100} \\ = \frac{90}{100} \\ = \frac{9}{10} [/tex]

Now,

The work left to be completed =

[tex]1 - \frac{9}{10} \\ = \frac{10}{10} - \frac{9}{10} \\ = \frac{1}{10} [/tex]

From this we know that,

Amber completes [tex]\frac{1}{10}[/tex] of the work in 3 days.

So,

Time taken by Amber to complete the whole work =

[tex]3 \times 10 \\ = 30 \: \: days[/tex]

↦ If Amber alone would do the whole work, he would take [tex]\boxed{30 \ \ days}[/tex] to complete it.

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ ツ

꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐

Answer:

B. (4,2)

Step-by-step explanation:

Answer:

B(4,2)

Step-by-step explanation:

as you can see that if want to find coordinates u should know that the position of x and y is (x,y). So u can that c on the x is lower than 5 so that u can say it is 4 and y is too far away from 5 also u it will be 2.

Answer:

G

Step-by-step explanation:

I did this on edge and it was right

Answer:

O  Point G

Step-by-step explanation:

A = -6

B = 8

To find the midpoint, calculate how much would it take for both points to have a value of zero.

-6 + ? = 0

-6 + 6 = 0

8 - ? = 0

8 - 8 = 0

so the midpoint will be about 7 (between 6 & 8)

Now which of the points best shows 7 units apart of AB

Answer: point G

Answer:

444

Step-by-step explanation:

3 [ ( 15 - 3 )^2 + 4 ]

= 3 [ ( 15 - 3 )^2 + 4 ]

= 3 [ ( 12 )^2 + 4 ]

= 3 [ 144 + 4 ]

= 3 [ 148 ]

= 444

Step-by-step explanation:

3(15-3)²+4

3(12)²+4

3×144+4

432+4

436 Answer

Answer:

[tex]10[/tex].

Step-by-step explanation:

See below for a proof of why all but the first digit of this [tex]N[/tex] must be "[tex]9[/tex]".

Taking that lemma as a fact, assume that there are [tex]x[/tex] digits in [tex]N[/tex] after the first digit, [tex]\text{A}[/tex]:

[tex]N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{$x$ digits}}}[/tex], where [tex]x[/tex] is a positive integer.

Sum of these digits:

[tex]\text{A} + 9\, x= 2021[/tex].

Since [tex]\text{A}[/tex] is a digit, it must be an integer between [tex]0[/tex] and [tex]9[/tex]. The only possible value that would ensure [tex]\text{A} + 9\, x= 2021[/tex] is [tex]\text{A} = 5[/tex] and [tex]x = 224[/tex].

Therefore:

[tex]N = \overline{5 \, \underbrace{9 \cdots 9}_{\text{$224$ digits}}}[/tex].

[tex]N + 1 = \overline{6 \, \underbrace{000 \cdots 000000}_{\text{$224$ digits}}}[/tex].

[tex]N + 2021 = 2020 + (N + 1) = \overline{6 \, \underbrace{000 \cdots 002020}_{\text{$224$ digits}}}[/tex].

Hence, the sum of the digits of [tex](N + 2021)[/tex] would be [tex]6 + 2 + 2 = 10[/tex].

Lemma: all digits of this [tex]N[/tex] other than the first digit must be "[tex]9[/tex]".

Proof:

The question assumes that [tex]N\![/tex] is the smallest positive integer whose sum of digits is [tex]2021[/tex]. Assume by contradiction that the claim is not true, such that at least one of the non-leading digits of [tex]N[/tex] is not "[tex]9[/tex]".

For example: [tex]N = \overline{(\text{A})\cdots (\text{P})(\text{B}) \cdots (\text{C})}[/tex], where [tex]\text{A}[/tex], [tex]\text{P}[/tex], [tex]\text{B}[/tex], and [tex]\text{C}[/tex] are digits. (It is easy to show that [tex]N[/tex] contains at least [tex]5[/tex] digits.) Assume that [tex]\text{B} \![/tex] is one of the non-leading non-"[tex]9[/tex]" digits.

Either of the following must be true:

If [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex], is a "[tex]0[/tex]", then let [tex]N^{\prime}[/tex] be [tex]N[/tex] with that "[tex]0\![/tex]" digit deleted: [tex]N^{\prime} :=\overline{(\text{A})\cdots (\text{B}) \cdots (\text{C})}[/tex].

The digits of [tex]N^{\prime}[/tex] would still add up to [tex]2021[/tex]:

[tex]\begin{aligned}& \text{A} + \cdots + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + 0 + \text{B} + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}[/tex].

However, with one fewer digit, [tex]N^{\prime} < N[/tex]. This observation would contradict the assumption that [tex]N\![/tex] is the smallest positive integer whose digits add up to [tex]2021\![/tex].

On the other hand, if [tex]\text{P}[/tex], the digit in front of [tex]\text{B}[/tex], is not "[tex]0[/tex]", then [tex](\text{P} - 1)[/tex] would still be a digit.

Since [tex]\text{B}[/tex] is not the digit [tex]9[/tex], [tex](\text{B} + 1)[/tex] would also be a digit.

let [tex]N^{\prime}[/tex] be [tex]N[/tex] with digit [tex]\text{P}[/tex] replaced with [tex](\text{P} - 1)[/tex], and [tex]\text{B}[/tex] replaced with [tex](\text{B} + 1)[/tex]: [tex]N^{\prime} :=\overline{(\text{A})\cdots (\text{P}-1) \, (\text{B} + 1) \cdots (\text{C})}[/tex].

The digits of [tex]N^{\prime}[/tex] would still add up to [tex]2021[/tex]:

[tex]\begin{aligned}& \text{A} + \cdots + (\text{P} - 1) + (\text{B} + 1) + \cdots + \text{C} \\ &= \text{A} + \cdots + \text{P} + \text{B} + \cdots + \text{C} \\ &= 2021\end{aligned}[/tex].

However, with a smaller digit in place of [tex]\text{P}[/tex], [tex]N^{\prime} < N[/tex]. This observation would also contradict the assumption that [tex]N\![/tex] is the smallest positive integer whose digits add up to [tex]2021\![/tex].

Either way, there would be a contradiction. Hence, the claim is verified: all digits of this [tex]N[/tex] other than the first digit must be "[tex]9[/tex]".

Therefore, [tex]N[/tex] would be in the form: [tex]N = \overline{\text{A} \, \underbrace{9 \cdots 9}_{\text{many digits}}}[/tex], where [tex]\text{A}[/tex], the leading digit, could also be [tex]9[/tex].

Hello!

4/18 = 6/27 ?

4 × 27 = 18 × 6

108 = 18 × 6

108 = 108 => 4/18 = 6/27

4/6 = 16/36 ?

4 × 36 = 6 × 16

144 = 6 × 16

144 ≠ 96 => 4/6 ≠ 16/36

3/4 = 9/12 ?

3 × 12 = 4 × 9

36 = 4 × 9

36 = 36 => 3/4 = 9/12

5/9 = 8/12 ?

5 × 12 = 9 × 8

60 = 9 × 8

60 ≠ 72 => 5/9 ≠ 8/12

Good luck! :)

X -Y =44 will represent the number of glasses of ice tea .

Answer:

C. $516

Step-by-step explanation:

His regular rate is $8 per hour.

Since it's 40 hours a week, it means from Monday to Friday his regular work time is 8 hours per day.

Thus, for the regular week work, he is to be paid;

40 × 8 = $320

Now, we are told he worked 9 hours each day from Monday through Friday.

This means that;

He worked 1 hour each day.

That is 5 hours extra from Monday to Friday.

He is paid one and one-half times the regular hourly rate.

Thus, for this 5 extra hours, he will be paid 1½ × 5 × 8 = $60

He works 6 hours on Saturday, and 4 hours on Sunday.

Thus;

For Saturday, he is also paid one and one-half of regular pay. Thus, he is due for;

1½ × 8 × 6 = $72

He is paid double the regular hourly pay for Sundays.

Thus, for 4 hours on Sunday, he is paid;

2 × 8 × 4 = $64

Total he is due = $320 + $60 + $72 + $64 = $516

Answer:

C)Ok i pick the point (18,4)

this point represents that if this person studied from 18 hours they got a GPA of 4.0

D) the chart below is the scatter plot

Hope This Helps!!!

Answer:

C. 40

Step-by-step explanation:

The smallest factor of 119 (other than 1) is 7.

28/7 = 4,

35/7 = 5,

40/7 = 5 5/7

63/7 = 9

So it is 40

Answer: B

Step-by-step explanation:

To find the values of x, we first need to write the function into an equation. We can derive 2 equations from the problem.

Equation 1: y=2|x+6|-4

Equation 2: y=6

Now, we can substitute.

2|x+6|-4=6

Let's solve for x.

2|x+6|-4=6        [add both sides by 4]

2|x+6|=10          [divide both sides by 2]

|x+6|=5              [subtract both sides by 6]

x=-1

Now that we know x=-1 is one of the solutions, we can eliminate C and D.

We know that the absolute value makes the number inside positive always. Therefore, let's solve for x with -5 instead.

|x+6|=-5              [subtract both sides by 6]

x=-11

Therefore, we know that B is the correct answer.

Answer:

you should make the picture more clear, i cant see the answer choices

Answer:

use pythogoream theorem that is,

Step-by-step explanation:

c*2=a*2+b*2

so in order to find a*2;

a*2=c*2-b*2

Answer:

nom nom

Step-by-step explanation:

nom nommy nom nom

Answer:

x - 2y + 8 = 0

Step-by-step explanation:

that is the procedure above

Answer:

(0,3) ; (2,3) ; (0,0) ; (3.5,0)

Step-by-step explanation:

Firstly, we have to plot all the giving inequalities as constraints

Not to forget, f(x) can be written as y

Kindly find the plot as an attachment

Upon plotting, we have the following vertices;

(0,3) ; (2,3) ; (0,0) ; (3.5,0)

Answer:

60

Step-by-step explanation:

5x + 10y = 800

y = 50

5x + 10(50) = 800

5x + 500 = 800

5x = 300

x = 60

Answer: 60

Step-by-step explanation:

If x = small vehicles and if y = large vehicles and y = 50, then substitute and solve for x. So,

5x + 10(50) = 800

5x + 500 = 800

5x = 300

x = 60

So, the number of small vehicles is 60.

Answer:

AM = 25, AC = 15, CM = 20

Step-by-step explanation:

The given parameters are;

In ΔACM, ∠C = 90°, [tex]\overline{CP}[/tex] ⊥ [tex]\overline{AM}[/tex], AP = 9, and PM = 16

[tex]\overline{AC}[/tex]² + [tex]\overline{CM}[/tex]² = [tex]\overline{AM}[/tex]²

[tex]\overline{AM}[/tex] = [tex]\overline{AP}[/tex] + PM = 9 + 16 = 25

[tex]\overline{AM}[/tex] = 25

[tex]\overline{AC}[/tex]² = [tex]\overline{AP}[/tex]² + [tex]\overline{CP}[/tex]² = 9² +  [tex]\overline{CP}[/tex]²

∴ [tex]\overline{AC}[/tex]² = 9² +  [tex]\overline{CP}[/tex]²

Similarly we get;

[tex]\overline{CM}[/tex]² = 16² + [tex]\overline{CP}[/tex]²

Therefore, we get;

[tex]\overline{AC}[/tex]² + [tex]\overline{CM}[/tex]² = 9² +  [tex]\overline{CP}[/tex]² + 16² + [tex]\overline{CP}[/tex]² = [tex]\overline{AM}[/tex]² = 25²

2·[tex]\overline{CP}[/tex]² = 25² - (9² + 16²) = 288

[tex]\overline{CP}[/tex]² = 288/2 = 144

[tex]\overline{CP}[/tex] = √144 = 12

From [tex]\overline{AC}[/tex]² = 9² +  [tex]\overline{CP}[/tex]², we get

[tex]\overline{AC}[/tex] = √(9² +  12²) = 15

[tex]\overline{AC}[/tex] = 15

From, [tex]\overline{CM}[/tex]² = 16² + [tex]\overline{CP}[/tex]², we get;

[tex]\overline{CM}[/tex] = √(16² + 12²) = 20

[tex]\overline{CM}[/tex] = 20.

Answer:

It's C

Step-by-step explanation:

Answer:

2026.75 (units) squared

Step-by-step explanation:

SA of a cone = πr(r+h2+r2)

SA = π(12)(12+√(40^2)+(12^2))

SA = 12π(12+√1744)

Put in calculator and get 2026.75

Answer:

ok ok ok ok ok ok ok

Step-by-step explanation:

Answer:

2 : 1

Step-by-step explanation:

Supplementary angles are two angles whose measures add up to 180°

If one of the angle = 120°

The other angle = sum of supplementary angle - one of the angle

= 180° - 120°

= 60°

The other angle = 60°

ratio of pair of supplementary aangle = 120° : 60°

= 120° / 60°

= 2/1

= 2 : 1

ratio of pair of supplementary aangle = 2 : 1

Answer:

[tex]x=7\text{ and } m\angle KLM = 34^\circ[/tex]

Step-by-step explanation:

We are given ethat KM and JN are parallel.

And we want to find the value of x.

Notice that ∠JKM and ∠LKM form a linear pair. Linear pairs total 180°. Therefore:

[tex]m\angle JKM + m\angle LKM = 180[/tex]

We know that ∠JKM measures (14x + 8). Substitute:

[tex](14x+8)+m\angle LKM =180[/tex]

Solve for ∠LKM:

[tex]m\angle LKM = 172-14x[/tex]

Next, since KM and JN are parallel, by the Corresponding Angles Theorem:

[tex]\angle JNM \cong \angle KML[/tex]

Since we know that ∠JNM measure (10x + 2), we can conclude that:

[tex]m\angle KML = 10x+2[/tex]

Next, recall that the three interior angles of a triangle must total 180°. Therefore:

[tex]m\angle KLM + m\angle LKM + m\angle KML = 180[/tex]

Substitute:

[tex](5x-1)+(172-14x)+(10x+2)=180[/tex]

Solve for x. Rewrite:

[tex](5x-14x+10x)+(-1+172+2)=180[/tex]

Combine like terms:

[tex](1x)+(173)=180[/tex]

Therefore:

[tex]x=7[/tex]

To find ∠KLM, substitute in 7 for x and evaluate. So:

[tex]m\angle KLM = 5(7) - 1 =34^\circ[/tex]