Need help solving homework

Answer:

n=5

6n^2-15=6(5)^2-15 -> add 15 to both sides

6n^2=6(5)^2 -> divide both sides by 6

n^2=5^2 -> sqr both sides

n=5

Notice that

13 - 9 = 4

17 - 13 = 4

so it's likely that each pair of consecutive terms in the sum differ by 4. This means the last term, 149, is equal to 9 plus some multiple of 4 :

149 = 9 + 4k

140 = 4k

k = 140/4

k = 35

This tells you there are 35 + 1 = 36 terms in the sum (since the first term is 9 plus 0 times 4, and the last term is 9 plus 35 times 4). Among the given options, only the first choice contains the same amount of terms.

Put another way, we have

[tex]\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=0}^{35} (9 + 4k)[/tex]

but if we make the sum start at k = 1, we need to replace every instance of k with k - 1, and accordingly adjust the upper limit in the sum.

[tex]\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k-1=0}^{35+1} (9 + 4(k-1))[/tex]

[tex]\displaystyle 9 + 13 + 17 + \cdots + 149 = \sum_{k=1}^{36} (5 + 4k)[/tex]

Answer:

56

Step-by-step explanation:

cross multiply

0.06z - 0.36 = 3

0.06z = 3.36

z = 56

Answer:

3x + 3

Step-by-step explanation:

yeah-ya........ right?

Answer:

-11, f(-5)    1

Step-by-step explanation:

The equilibrium price of the commodity at the given demand and supply is 1,000.

The given parameters:

The equilibrium price of the commodity is calculated as follows;

At equilibrium price, the demand curve and supply curve cross each other.

[tex]QS= QD\\\\15,000 +2P = 20,000 - 3P\\\\2P+3P = 20,000 - 15,000\\\\5P = 5,000[/tex]

[tex]P = \frac{5,000}{5} \\\\P = 1,000[/tex]

Thus, the equilibrium price of the commodity is 1,000.

The complete question is below;

Determine the equilibrium price, if QD = 20000 -3P  and QS = 15000 +2P.

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Answer:

Step-by-step explanation:

2 x 10¹² / 5 x 10⁶ = 2/5 x 10⁽¹²⁻⁶⁾ = 0.4 x 10⁶ = 4 x 10⁵

Answer:

x^2 - 9

Step-by-step explanation:

Multiply parenthesis

x × x = x^2

x × -3 = -3x

3 × x = 3x

3 × -3 = -9

x^2 - 3x + 3x - 9

= x2 - 9

Answer:

20.50609665

around 20.506

Answer:

1: Out: 5, 8, 11, 14

2: Out: 2, 7, 10, 11

3: Out: 11  Rule: Add 4

4: Out: 6  Rule: Add 4

Answer:

360 people

Step-by-step explanation:

[tex]\frac{ 210}{525}*900[/tex][tex]=360[/tex]

Hope this helped! :)

Answer:

From the time Ms. Cruz purchased a stock till closing time that day, the value of the stock changed by -27.2 points. During that time, the stock value

changed at a steady rate of -6.4 points per hour.

Answer:

C.

Ax + 1 = x A + 1

Bx + 2 = x B + 2

Cx + 3 = x C + 3

A = x C + 3

B = x C + 3

C = x C + 3

I'm not really sure about the answer but i hope it helps

Here is one way to show that the left side is identical to the right side.

[tex]\sec(x)-\tan(x)\sin(x) = \cos(x)\\\\\\\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}\sin(x) = \cos(x)\\\\\\\frac{1}{\cos(x)}-\frac{\sin^2(x)}{\cos(x)} = \cos(x)\\\\\\\frac{1-\sin^2(x)}{\cos(x)} = \cos(x)\\\\\\\frac{\cos^2(x)}{\cos(x)} = \cos(x)\\\\\\\cos(x)=\cos(x)\\\\\\[/tex]

Throughout the entire process, the right hand side stayed the same. When working on identities, it's important to keep one side the same while you transform the other side.

Despite the two sides being identical when simplified, there is the issue of cos(x) being zero in the denominator on the left hand side. Be sure to account for this when forming the domain. No such issue occurs on the right hand side. For instance, the value x = pi/2 leads to a division by zero error when in radian mode. So if I was your teacher, I would revise the "for any value of x" and replace it with something along the lines of "for any x value in the domain".

Answer:

Katrina ran 1 mile 3 times! I hope this helps you!

Answer:14mm

Step-by-step explanation:

Answer:

Adult males are flame-orange and black, with a solid-black head and one white bar on their black wings. Females and immature males are yellow-orange on the breast, grayish on the head and back, with two bold white wing bars

Answer:

It is 18 feet

Step-by-step explanation:

The remaining spoon of ribbon will be; 22 feet.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

We have been given that a spool of ribbon comes with 24 feet of ribbon If 2 yards of ribbon are cut off, then

WE need to find the remaining spoon of ribbon after excluding the 2 yards of ribbon.

The remaining spool can be calculated by the difference;

24 feet of ribbon - 2 yards of ribbon

24 - 2 = 22

Therefore, the remaining spoon of ribbon will be; 22 feet.

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Answer:

The area of real patio is 50m².

Step-by-step explanation:

Given,

Length of patio = 3 cm

Width of patio = 6 cm

According to scale =

3 cm = 5m

1 cm = \frac{5}{3}\ m35 m

6 cm = \frac{5}{3}*6 = 10\ m35∗6=10 m

Real length of patio = 5 m

Real width of patio = 10 m

Area = Length*width

\begin{gathered}Area=5*10\\Area=50\ m^2\end{gathered}

Area=5∗10Area=50 m2

The area of real patio is 50m².

Answer:

[tex]y=\frac{1}{2} x+9.[/tex]

Step-by-step explanation:

1) all the perpendicular lines are: 2x-4y=C, where C - unknown number;

2) if to substitute the coordinates (-8;5) into 2x-4y=C, then C=2*(-8)-4*5= -36;

3) if C= -36, then the required equation is: 2x-4y=-36 or y=1/2 x +9.

like my answer and brainliest

Answer:

B.

Step-by-step explanation:

Adding 0 to 6 gives the same value (6).

Zero is called the Identity element of Addition,

Answer:

a=3

Step-by-step explanation:

you subtract 2 from both sides of the equation you simplify it and then you divide both sides of the equation by the same term and then simplify it again

Answer:

a=3

Step-by-step explanation:

The amount Howie need to pay back at the end of one year, including the interest is $26,000

Given:

Interest rate = 4%

Cost of car = $25,000

Time = 1 year

Amount of interest = Principal × Rate × Time

= 25,000 × 4% × 1

= 25,000 × 0.04 × 1

= $1,000

Amount to be paid = Amount of interest + Cost of car

= 1,000 + 25,000

= $26,000

Therefore, the amount Howie need to pay back at the end of one year, including the interest is $26,000

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The amount that Howie would need to pay back at the end of one year, including the interest for a car loan of $25,000 is $26,000.

Simple interest is a finance charge that does not compound.  It is calculated on the principal portion of a loan only.

The formula for computing simple interest is Principal x APR x Number of years.

Car loan = $25,000

Simple interest rate = 4%

Period of loan = 1 year

Principal + Interest for 1 year = $26,000 ($25,000 x 4% x 1).

Thus, the amount that Howie would need to pay back at the end of one year, including the interest for a car loan of $25,000 is $26,000.

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Answer:

0.0075 centimeters long

Step-by-step explanation:

An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.

L x 100 = 0.75 cm

The actual length of the specimen = L

The actual length of the specimen is 0.0075 cm

An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.

Example: 2 + 3x + 4y = 7 is an expression.

We have,

Seen through the microscope, a specimen is 0.75 cm long.

The Microscope enlarges objects to 100 times their actual size.

This means,

Let the actual length of the specimen = L

Now,

L x 100 = 0.75 cm

The actual length of the specimen.

L x 100 = 0.75 cm

L = 0.75 / 100

L = 75/10000

L = 0.0075 cm

Thus,

The actual length of the specimen is 0.0075 cm

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Answer:

C    [tex] \dfrac{9a^{12}}{b^3} [/tex]

Step-by-step explanation:

54/6 = 9

a^8 / a^-4 = a^{8 - (-4)} = a^12

b^2 / b^5 = b^-3 = 1/b^3

Answer:  C    [tex] \dfrac{9a^{12}}{b^3} [/tex]

Answer:

Option C

Step-by-step explanation:

[tex] \sf \dfrac{54 {a}^{8} {b}^{2} }{6 {a}^{ - 4} {b}^{5} } [/tex]

We know that

[tex] \sf \: \frac{ {a}^{m} }{a {}^{n} } = a {}^{m - n} [/tex]

[tex] \sf 9 \dfrac{a {}^{8} {b}^{2} }{ {a}^{ - 4} {b}^{5} } [/tex]

[tex] \sf \: 9 \dfrac{ {a}^{8 - ( - 4)} b {}^{(2 - 5)} }{1} [/tex]

[tex] \sf \: 9 \dfrac{ {a}^{12}b {}^{ - 3} }{1} [/tex]

[tex] \sf \: \dfrac{9 {a}^{12} }{ {b}^{3} } [/tex]