MATHEMATICS HIGH SCHOOL

Which equation has an a-value of 1, a b-value of –3, and a c-value of –5?0 = –3x – 5 + x2
0 = x – 3 – 5x2
0 = 3x – 5 – x2
0 = –3x + 5 – x2

Answers

Answer 1
Answer:

Answer:

Option (a) is correct.

For given values a = 1 , b = -3 , c = -5 the quadratic equation is  0=-3x-5 + x^2

Step-by-step explanation:

Given : a = 1 , b = -3 , c = -5

We have to write the quadratic equation having a = 1 , b = -3 , c = -5 and choose the correct option.

The standard form of quadratic equation is ax^2+bx+c=0 , where a, b, c are constant integers.

Given : a = 1 , b = -3 , c = -5

Then Substitute, we get,

x^2-3x-5=0  

Thus, the obtained quadratic equation is same as option (a) 0=-3x-5 + x^2

Thus, For given values a = 1 , b = -3 , c = -5 the quadratic equation is0=-3x-5 + x^2 
Answer 2
Answer:

Answer:

a

Step-by-step explanation:

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Choose the correct product of (2x − 4)2.4x2 − 16 4x2 − 16x + 16 4x2 + 16 4x2 + 16x + 16

There are 81 new houses being built in a neighborhood. Last month, 1 over 3 of them were sold. This month, 2 over 3 of the remaining houses were sold. How many houses are left to be sold?

Answers

81 houses total.

(1)/(3) of 81 (27) sold last month.

81 - 27 = 54 houses remaining.

(2)/(3) of 54 (36) sold this month.

54 - 36 = 18 houses remaining.

There are 18 houses left to be sold.

Final answer:

There were 81 new houses. Initially, 1/3 of them were sold, leaving 54 houses. Later on, 2/3 of the remaining houses were sold, leaving a total of 18 houses left to be sold.

Explanation:

The subject of this question is Mathematics and it appears to be suitable for a Middle School level. The problem is about understanding fractions applied to a real-life situation, specifically the sales of new houses in a neighborhood.

Let's begin by working out how many of the 81 houses were sold last month. Because the question states that '1 over 3' were sold, this equates to 81 divided by 3, which is 27 houses. We then deduct this number from the total of 81 to find the remaining houses. So, 81 - 27= 54 houses remained.

In this month, '2 over 3' of the remaining houses were sold. So, we calculate 2/3 of 54. Multiply 54 by 2 and then divide this by 3 to get 36 houses being sold this month.

Subtracting these 36 houses from the remaining 54 (i.e., 54 - 36), we find that there are 18 houses left to be sold.


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Given the geometric series:16+2+1/4... find the sum of the geometric progression up to the 6th term​

Answers

The sum of the geometric progression up to the 6th term is 512/7

To find the sum of a geometric series, you can use the formula for the sum of a finite geometric series:

\[S_n = (a(1 - r^n))/(1 - r)\]

Where:

- Sₙ is the sum of the first n terms of the series.

- a is the first term of the series.

- r is the common ratio.

- n is the number of terms in the series.

In your case, you have the geometric series: 16, 2, 1/4, ...

1. Identify the values for the formula:

  - The first term (a) is 16.

  - The common ratio (r) is found by dividing the second term by the first term: 2/16 = 1/8.

  - You want to find the sum of the first 6 terms (n = 6).

2. Plug these values into the formula and calculate S₆:

S₆ = (16(1 - (1/8)⁶))/(1 - 1/8)

Now, calculate the individual terms in the formula:

S₆ = (16(1 - 1/262144))/(7/8)

S₆ = (16(262143/262144))/(7/8)

S₆ = ((4194288/262144)/(7/8)

Now, perform the division:

S₆ = (4194288/262144) \* (8)/(7)

S₆ = 64 * 8/7

Now, multiply:

S₆ = 512/7

So, the sum of the geometric progression up to the 6th term is 512/7

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How many grams in an ounce

Answers

Answer is provided in the image attached.
In 1 ounce, there is approximately 28.34 grams.

Find the diagonal of a square whose area is 64 M2

Answers

Answer:

Square area =c²

64=c²

c=√64 = 8m

we calculate the diagonal

d²=8²+8²

d²=128

d= √128 = 8√2 exact value

Step-by-step explanation:

Someone please help??

Answers

4 is the correct answer as 4th root of 1296 is 6

The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical. The Sun is located at a focus of the ellipse. The perihelion is the point in a planet’s orbit that is closest to the Sun. So, it is the endpoint of the major axis that is closest to the Sun.

The aphelion is the point in the planet’s orbit that is furthest from the Sun. So, it is the endpoint of the major axis that is furthest from the Sun.

The closest Mercury comes to the Sun is about 46 million miles. The farthest Mercury travels from the Sun is about 70 million miles.


1. What is the distance between the perihelion and the aphelion?


2. What is the distance from the center of Mercury’s elliptical orbit and the Sun?


3. Write the equation of the elliptical orbit of Mercury, where the major axis runs horizontally. Allow a and b to be measured in millions of miles. Use the origin as the center of the

Answers

Answer:

(1) 83.764 million miles

(2) 52.766 million miles

(3) (x^2)/(4900)+(y^2)/(2116)=1.

Step-by-step explanation:

Let the origin C(0,0) be the center of the elliptical path as shown in the figure, where the location of the sun is at one of the two foci, say f.

The standard equation of the ellipse having the center at the origin is

(x^2)/(a^2)+(y^2)/(b^2)=1\;\cdots(i)

where a and b are the semi-axes of the ellipse along the x-axis and y-axis respectively.

Let the points P and A represent the points of perihelion (nearest to the sun) and the aphelion (farthest to the sun) of the closest planet Mercury.

Given that,

CP=46 million miles and

CA=70 million miles.

So, CP=b is the semi-minor axis and CA=a is the semi-major axis.

Let the distances on the axes are in millions of miles. So, the coordinates of the point P and A are P(0,46) and A(70,0) respectively.

(1) From the distance formula, the distance between the perihelion and the aphelion is

PA=√((0-70)^2+(46-0)^2)=83.764 million miles.

(2) Location of the Sun is at focus, f, of the elliptical path.

From the standard relation, the distance of the focus from the center of the ellipse, c, is

c=ae\;\cdots(ii)

where a and e are the semi-major axis and the eccentricity of the ellipse.

The eccentricity of the ellipse is

e=\sqrt{1-(b^2)/(a^2)}

\Rightarrow e=\sqrt{1-(46^2)/(70^2)}=0.7538.

Hence, from the equation (i) the distance of the Sun from the center of the elliptical path of the Mercury is

c=70*0.7538=52.766 million miles.

(3) From the equation (i), the equation of the elliptical orbit of Mercury is

(x^2)/(70^2)+(46^2)/(b^2)=1

\Rightarrow (x^2)/(4900)+(y^2)/(2116)=1.
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