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2 years ago

A line has a slope of zero Which of the following points could this line pass through

Mathematics

1 answer:

Rina8888 [55]2 years ago

3 0

Answer:

Step-by-step explanation:

m(slope) = y2 - y1 / x2 - x1

-6 - - 6 / 7 - 3

= 0 / 4 = 0

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Eli, Daphne, Celeste, and Thaddeus ran for school president. The graph shows how the students in their school voted.

andrew11 [14]

Eli received the most votes

Step-by-step explanation:

Given percentages are:

Eli: 30%

Daphne: 27%

Celeste: 25%

Thaddeus: 18%

When a circle chart is used, the sum of all percentages is 100.

In case of finding the largest between the portions, the part with the greater percentage has the greater share in the chart.

According to given percentages, Eli has the highest percentage, which means that Eli received the most votes.

Hence,

Eli received the most votes

Keywords: Circle chart, percentages

Learn more about circle chart at:

#LearnwithBrainly

4 0

2 years ago

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(5x10^3)x(9x10^7) can someone please answer this for me.

Anna11 [10]

4.5 times 10^10

1. Multiply 5times ten to the third

2. Multiply 9 times ten raised to the seventh

Then multiply them together
5000 times 90000000

5 0

2 years ago

AWNERSSS FOR ALLL PLSSSSSSSS 67 pointssss

melomori [17]

Step-by-step explanation:

4) x=(z-y)/2

8) b=(d)/ac

9) x=(-z+4)/y

10) a=(c-3)b

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2 years ago

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60 % of 80 ASAP pls bruuh

Sati [7]

Answer:

Step-by-step explanation:

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2 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

2 years ago