INCOME; you earn a weekly salary of $200 plus a 3% commission on the total value of the sales made in the week. This week, your sales total $2000. What are your total earnings for the week?

#1 identify the percent, bass, and part of the base in the question

The values of the zero are 1 and 3 if the quadratic function is m(x) = x² - 4x + 3 after solving algebraically.

Any equation of the form [tex]\rm ax^2+bx+c=0[/tex]  where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

We have quadratic function:

m(x) = x² -4x + 3

m(x) = x² -x - 3x + 3

m(x) = x(x -1) -3(x - 1)

m(x) = (x - 1)(x - 3)

x -1 = 0 or x -3 = 0

x = 1  or  x = 3

Thus, the values of the zero are 1 and 3 if the quadratic function is m(x) = x² - 4x + 3 after solving algebraically.

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The zeros of the quadratic function m(x) = x² - 4x + 3 are found using the quadratic formula, yielding two solutions, x = 1 and x = 3.

To find the zeros of the quadratic function m(x) = x² - 4x + 3, we first note that it is already in standard form, with a = 1, b = -4, and c = 3. The quadratic formula, which states if ax² + bx + c = 0, then x = (-b ± √(b² - 4ac)) / (2a), can be used to find the solutions. Substituting the values, we get:

x = (4 ± √((-4)² - 4(1)(3))) / 2(1)

x = (4 ± √(16 - 12)) / 2

x = (4 ± √4) / 2

x = (4 ± 2) / 2

This results in two solutions:

x = (4 + 2) / 2 = 6 / 2 = 3

x = (4 - 2) / 2 = 2 / 2 = 1

Therefore, the zeros of the function m(x) are 1 and 3.

Final answer:

The store sold 1 and 2/5 more cases of juice on Friday than on Saturday.

Explanation:

To determine how many more cases of juice the store sold on Friday than on Saturday, we need to subtract the number of cases sold on Saturday from the number of cases sold on Friday.

Friday's sales: 6 and 1/5 cases
Saturday's sales: 4 and 4/5 cases

We convert these mixed numbers to improper fractions to make the subtraction easier:

Now we subtract the two fractions:

(31/5) – (24/5) = 7/5 cases

Convert the improper fraction back to a mixed number:

7/5 = 1 and 2/5 cases

So, the store sold 1 and 2/5 more cases of juice on Friday than on Saturday.

Each peach costs $0.56 and the equations used to solve that is [tex]4x= 2.24[/tex] where x is the cost of 1 peach.

Step-by-step explanation:

Step 1:

It is given that all the peaches weigh the same and that 4 peaches weigh 1 pound.

Assume that each peach costs x.

So the cost for 4 peaches is given by [tex]4x.[/tex]

Step 2:

The cost of 1 pound of peaches is given as $2.24.

1 pound of peaches[tex]=4x.[/tex]

So [tex]4x= 2.24, x = \frac{2.24}{4}.[/tex]

[tex]x = 0.56.[/tex]

So each peach costs $0.56.

Answer:

32

Step-by-step explanation:

let a = 5

a² + 7 | Original equation

5² + 7 | Substitution of a as 5

25 + 7 | 5 to the power of 2 is 25, as 5*5 = 25.

32       | 25 + 7 = 32

Answer:

2-2-2

History

The 2-2-2 configuration appears to have been developed by Robert Stephenson and Company in 1834, as an enlargement of their 2-2-0 Planet configuration, offering more stability and a larger firebox. The new type became known as Stephenson's Patentee locomotive.[1] Adler, the first successful locomotive to operate in Germany, was a Patentee supplied by Robert Stephenson and company in component form in December, 1835 was one of the earliest examples. Other examples were exported to the Netherlands, Russia and Italy.[2] By 1838 the 2-2-2 had become the standard passenger design by Robert Stephenson and Company.[3]

Eighteen of the first nineteen locomotives ordered by Isambard Kingdom Brunel for the opening of the Great Western Railway in 1837/8 were of the 2-2-2 type.[4] These included six 2-2-2 locomotives built by Charles Tayleur at his Vulcan Foundry. Also in 1837 the successful North Star broad gauge locomotive was delivered to the Great Western Railway by Stephenson, becoming the first of a class of twelve locomotives by 1841.

Great Western Railway North Star at Swindon

Later UK developments

Sharp, Roberts and Company constructed more than 600 2-2-2 locomotives between 1837 and 1857. Ten of these supplied to the Grand Junction Railway became the basis of Alexander Allan's successful designs for the railway from 1845 (the first of which, formerly named Columbine, is preserved). J. & G. Rennie supplied 2-2-2 locomotives to the London and Croydon Railway from 1838 and the London and Brighton Railway in 1840.[5] Arend ("eagle") was one of the two first steam locomotives in the Netherlands, built by R. B. Longridge and Company of Bedlington, Northumberland in 1839.

The Great Western Railway continued to order both broad gauge and standard gauge locomotives on the railway, including the Firefly and Sun classes (1840–42), which were enlarged versions of North Star. Bury, Curtis, and Kennedy supplied six 2-2-2 locomotives to the Bristol and Gloucester Railway in 1844, and fourteen to the Great Southern and Western Railway in Ireland in 1848, (the last of these has been preserved at Cork Kent railway station.

The original "Jenny Lind" locomotive, 1847.

The Jenny Lind locomotive, designed by David Joy and built in 1847 for the London Brighton and South Coast Railway by the E.B.Wilson and Company of Leeds, became the basis of hundreds of similar passenger locomotives built during the 1840s and 1850s by this and other manufacturers for UK railways. The London & North Western Railway Cornwall locomotive was designed at Crewe Works as a 4-2-2 by Francis Trevithick in 1847, but was rebuilt as a 2-2-2 in 1858.

Although by the 1860s the 2-2-2 configuration was beginning to be superseded by the 2-4-0 type with better adhesion, the invention of steam sanding gave 2-2-2 singles a new lease of life, and they continued to be built until the 1890s. Notable late examples include William Stroudley's singles of 1874-1880, William Dean's 157 class of 1878-79,[6] and his 3001 class (1891–92),[7][8] both for the Great Western Railway. James Holden of the Great Eastern Railway created some 2-2-2 singles in 1889 by removing the coupling rod from a 2-4-0.

Belgium

Replica of 'Le Belge' 1835

The first steam railway locomotive built in Belgium in 1835, and was built by John Cockerill under license to a design by Robert Stephenson & Co. It was built for use on the first main line on the European mainland, the Brussels-Mechelen line.[9] A replica was built at the workshops of Boissellerie Cognaut for the 150th anniversary of the formation of Belgium.[10]

Italy

Two 2-2-2 locomotives were imported from Longridge and Co of Bedlington Ironworks England for the Naples–Portici railway in 1839 named Bayard and Vesuvio. A replica of 'Bayard is at the Naples Railway Museum.[11]

Germany

Most of the earliest locomotives to operate in what is now Germany before the mid-1840s were 2-2-2s delivered by UK manufacturers. However, by 1839 the type was also being built locally see List of Bavarian locomotives and railbuses. The Pegasus of 1839 was the first locomotive to be built by the Sächsische Maschinenbau-Compagnie in Chemnitz. August Borsig and Company manufactured Beuth in 1843 which was highly successful; its valve design became de facto standard for locomotives for decades to come.[12] By 1846 he had manufactured more than a hundred similar locomotives. Both the Leipzig-Dresden Railway and Royal Bavarian State Railways (Königlich Bayerische Staatsbahn) built several 2-2-2 classes 1841-1859. Similarly, the Grand Duchy of Mecklenburg Friedrich-Franz Railway grouped various 2-2-2 steam locomotives procured from German manufacturers between 1848 and 1863 into its Mecklenburg I class.

Answer:

[tex]r=3.93\%[/tex]

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

[tex]t=18\ years\\ P=x\\A=2x\\ r=?\\n=1[/tex]  

substitute in the formula above

[tex]2x=x(1+\frac{r}{1})^{1*18}[/tex]  

[tex]2=(1+r})^{18}[/tex]  

Elevated both sides to 1/18

[tex]2^{\frac{1}{18}} =1+r[/tex]

[tex]r=2^{\frac{1}{18}} -1[/tex]

[tex]r=0.0393[/tex]

convert to percentage

Multiply by 100

[tex]r=3.93\%[/tex]

Answer:

[tex]H( - 2) = - 7 \frac{1}{5} [/tex]

Step-by-step explanation:

We want to evaluate the equation

[tex]H(y) = \frac{9}{5} \times 2y[/tex]

at y=-2

We just have to substitute to obtain;

[tex]H( - 2) = \frac{9}{5} \times 2 \times - 2[/tex]

We now multiply to get:

[tex]H( - 2) = \frac{9}{5} \times - 4[/tex]

This is the same as:

[tex]H( - 2) = \frac{9}{5} \times \frac{ - 4}{1} [/tex]

We multiply the numerators separately and denominators separately to get:

[tex]H( - 2) = \frac{9 \times - 4}{5 \times 1} [/tex]

[tex]H( - 2) = \frac{ - 36}{5} [/tex]

This implies that

[tex]H( - 2) = - 7 \frac{1}{5} [/tex]

Answer:

Option D, x = 40

Step-by-step explanation:

Since x and (4x - 20) are supplementary, that means they add up to 180 degrees.

Step 1:  Make an equation

x + (4x - 20) = 180

Step 2:  Combine like terms

x + 4x - 20 = 180

5x - 20 = 180

Step 3:  Add 20 to both sides

5x - 20 + 20 = 180 + 20

5x = 200

Step 4:  Divide both sides by 5

5x / 5 = 200 / 5

x = 40

Answer:  Option D, x = 40

Answer:

2n - 5 = n^2 (or n squared)

The equation that could be used to properly solve for n is [tex]n^2 - 2n - 5 = 0[/tex]

To properly solve for the number, n, based on the given information, you can use the equation:

[tex]2n = n^2 - 5[/tex]

This equation represents the statement "Twice a number, n, is 5 less than the number squared." It sets the twice of the number (2n) equal to the number squared minus 5 (n^2 - 5).

To solve for n, you can rearrange the equation as follows:

[tex]n^2 - 2n - 5 = 0[/tex]

You can then solve this quadratic equation for n using methods such as factoring, completing the square, or the quadratic formula.

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Answer:

y=-2

Step-by-step explanation:

Hello : let  A(-3,-4)    B(6,2)

the slope is :   (YB - YA)/(XB -XA)

(2+4)/(6+3)  = 6/9 = 2/3

an equation is the line is : y = ax+b        a is a slope

y = (2/3)x+b

but this line passes by (6;2)

so : 2 = (2/3)(6)+b

b = -2

the equation is : y = (2/3)x-2

the the Y intercept of this line when : x= 0   so : y = (2/3)(0)-2

y=-2

Answer:

Step-by-step explanation:

Answer:

x=√y+2 + 6

Step-by-step explanation:

y=(x-6)^2+2

Substrate 2 from both sides

y-2=(x-6)^2

Square root both sides

√y-6=x-6

Add 6 to both sides

√y-6 +6= x

Therefore,x=√y-6 + 6

Answer:

3bzycukznwvy knsheuje6d9qk3vhdurkgyvuf6dyguts bl es yvhuv

Answer:

g = 20

Step-by-step explanation:

−2 (g − 13) + 15 = 1

Step 1: Simplify both sides of the equation.

−2 (g−13) + 15 = 1

(−2) (g) + (−2) (−13) + 15 = 1 (Distribute)

−2g + 26 + 15 = 1

(−2g) + (26 + 15) = 1 (Combine Like Terms)

−2g + 41 = 1

−2g + 41 = 1

Step 2: Subtract 41 from both sides.

−2g + 41 − 41 = 1 − 41

−2g = −40

Step 3: Divide both sides by -2.

−2g/−2 = −40/−2

g = 20

Answer:

[tex]y=\frac{1}{4}x[/tex]    [tex]x=4y[/tex]

Step-by-step explanation:

Assuming

y = inches of rain

x = hours (The x latitude is usually used for time)

The scale factor of the dilation from ABCD to A′B′C′D′ is 3.

Step-by-step explanation:

Step 1:

In the pre-image ABCD, the length of one of the sides is given as 14 units.

For the other shape A′B′C′D′, the same side as the previous shape is given as 8 units.

Step 2:

To determine the scale factor, we divide the measurement after scaling by the same measurement before scaling.

In this case, it is the given length of the sides CD and C′D′.

So the scale factor = [tex]\frac{C^{1} D^{1} }{CD} = \frac{8}{14} = \frac{4}{7} .[/tex]

So the shape ABCD is dilated by a scale factor of [tex]\frac{4}{7}[/tex] to produce the shape A′B′C′D′.

Answer: The answer is 4/7

Step-by-step explanation: the other persons work was correct!! so i got 100%

Answer:

y=mx+b broken down is in the step by step explanation.

Step-by-step explanation:

y=mx+b is the y intercept formula.it is mostly used for slopes.

y is usually equal to the total,or final answer.it is the start.

m is equal to the slope.in the formula.it can also be used as the rise over run,which means the top number is the amount of units you go up for rise,and run is the bottom number which is the amount of units you move to the right.if it is negative,you move the opposite ways.if it is a single number,the bottom number can be 1.

x is equal to the number you multiply m,or the slope by.

b is equal to where you start on the vertical line in the graph.a way to remember it is horizontal is like a horizon,it’s flat.

Answer:

A. When x = 3, both expressions have a value of 30.

Step-by-step explanation:

The first expression is 4(x+3)+2x

The second expression is 6(x+2).

When we substitute x=3 into the first equation, we get:

[tex]4(3 + 3) + 2(3) = 4 \times 6 + 6 = 24 + 6 = 30[/tex]

When we substitute x=3 into the second expression we get:

[tex]6(3 + 2) = 6(5) = 30[/tex]

This proves that the two expressions are equivalent.

The correct choice is A

Answer: A

Step-by-step explanation:

Hope this helps can i have brainly-est please

Step-by-step explanation:

Total area of merged figure = area of vertical rectangle + area of horizontal rectangle.

[tex] \therefore \: 3x(2x - 7) + (3x + 4)x = 85 \\ \\ \therefore \: 6 {x}^{2} - 21x+ 3 {x}^{2} + 4x = 85 \\ \\ \therefore \: 6 {x}^{2} + 3 {x}^{2} - 21x + 4x - 85 = 0 \\ \\ \huge \red {\boxed {\therefore \: 9 {x}^{2} - 17x - 85 = 0}} \\ hence \: proved.[/tex]

Answer:

Step-by-step explanation:

(3x + 4)(x) + (2x - 7)(3x) = 85

3x² + 4x + 6x² - 21x = 85

9x² - 17x - 85 = 0

Answer:

This is a statement not a question

Here we need to solve some expressions. Remember the following rules:

[tex](+)(+)=+ \\ \\ (-)(-)=+ \\ \\ (+)(-)=- \\ \\ (-)(+)=-[/tex]

Then, from the left side:

[tex]5-2\cdot 7=5-14=\boxed{-9} \\ \\ 5-2\cdot 4=5-8=\boxed{-3} \\ \\ 5-2\cdot 0=5-0=\boxed{5} \\ \\ 5-2\cdot -3=5+6=\boxed{11} \\ \\ 5-2\cdot 10=5-20=\boxed{-15} \\ \\ 5-2\cdot -8=5+16=\boxed{21} \\ \\[/tex]

from the right side:

[tex]-(-4)+3=4+3=\boxed{7} \\ \\ -(6)+3=-6+3=\boxed{-3} \\ \\ -(-8)+3=8+3=\boxed{11} \\ \\ -(-5)+3=5+3=\boxed{8} \\ \\ -(2)+3=-2+3=\boxed{1} \\ \\ -(15)+3=-15+3=\boxed{-12} \\ \\ -(-6)+3=6+3=\boxed{9}[/tex]

Answer: 241

Step-by-step explanation:

This is a simple problem when using Pascal's triangle. First you look at 1 - 3, what is the difference in these 2 numbers? Its 2, then you look at 3 and 11, the difference is 8, then 11 and 35, the difference is 24. If you keep doing this you get; 2 8 24 64 142. Next you find out how these numbers are different. In the end you keep doing this until you get to the triangles point or the differences converge.

                                          1      3      11      35     99      241

                                             2      8     24       64    142

                                                6      16      40      78

                                                     10     24      38

                                                          14       14

In the end we get 14 but there is no difference in the numbers so we cant go any farther. Hope this helps!

Answer:

51

Step-by-step explanation:

Answer:

51°

Step-by-step explanation:

90 - 39 = 51

Answer: B) line reflection

The orientation will flip with a reflection. So if points A,B,C are going clockwise for instance, then A', B', and C' will go counterclockwise.

Answer:

5.04 inches

Step-by-step explanation:

because the total rainfall in March was 3.6 inches and if the total rainfall in April was 1.4 times as much then you would take the total rainfall in March which was 3.6 inches and multiply that by 1.4 because it says the amount in April was 1.4 times as much.

The total rainfall in March was 3.6 inches. In April, the total rainfall was 1.4 times as much. The total rainfall in April is calculated by multiplying the March rainfall of 3.6 inches by 1.4, resulting in a total April rainfall of 5.04 inches.

The total rainfall in March was 3.6 inches. In April, the total rainfall was 1.4 times as much. To calculate the total rainfall in April, you multiply the March rainfall by 1.4. So, the calculation you'd do is 3.6 inches (March rain) times 1.4. This gives you 5.04 inches of rainfall in April. So, the total rainfall in April was 5.04 inches.

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Answer:

So the answer will be 2 2/4

Step-by-step explanation:

If you need me to explain more please tell me.Hope it helps.

Jenna is left with 2 1/2 meters of ribbon for future projects.

Jenna bought a spool of ribbon that was 5 3/4 meters long and used 3 1/4 meters of it for a craft project. To calculate how much ribbon she has left, we need to subtract the length of the ribbon used from the total length of the ribbon.

Convert mixed numbers into improper fractions: 5 3/4 = (5 × 4) + 3 = 23/4 and 3 1/4 = (3 × 4) + 1 = 13/4.

Subtract the second fraction from the first: 23/4 - 13/4 = (23 - 13)/4 = 10/4.

Simplify the result: 10/4 = 2 2/4 = 2 1/2 meters.

Answer: $54

Step-by-step explanation:

Original price = $60

10% discount= 10/100 × 60

= 0.01 × 60 = $6

Amount cassie paid for the sweater is original price - discount

=$60 - 6

= $54

I hope this helps

Answer:

54$

Step-by-step explanation:

Answer:

hope it helps you see the attachment for further information

Answer:

See below

Step-by-step explanation:

Let's check:

34*1/2 does equal 17, so Yes

51*2/3 does equal 34, so Yes

34*3/8 does not equal 12, so No

300*1/9 does not equal 34, so No

Option D: Graph C is the solution to the system of linear inequalities.

Explanation:

The system of linear inequalities are [tex]y \geq 2 x+1[/tex] and [tex]y<-2 x-3[/tex]

Let us graph the inequality [tex]y \geq 2 x+1[/tex]

Let us substitute the coordinate [tex](0,0)[/tex] in the inequality [tex]y \geq 2 x+1[/tex], we get,

[tex]0 \geq 2 (0)+1[/tex]

[tex]0 \geq1[/tex]

This is false. Hence, the solution does not contain the coordinate [tex](0,0)[/tex]

Thus, let us shade the upper part of the line in the graph.

Similarly, let us substitute the coordinate [tex](0,0)[/tex] in the inequality [tex]y<-2 x-3[/tex], we get,

[tex]0<-2 (0)-3[/tex]

[tex]0<-3[/tex]

This is false. Hence, the solution does not contain the coordinate [tex](0,0)[/tex]

Thus, let us shade the lower part of the line in the graph.

The solution to the system of linear inequality is the intersection of the two regions.

Thus, the solution of the inequality is Graph C.

Hence, Option D is the correct answer.