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2 years ago

23 - C<6 I cannot find the answer can someone help me?

Mathematics

1 answer:

Blababa [14]2 years ago

5 0

Answer:

C > 17

Step-by-step explanation:

23 - C < 6

-23 -23

-----------------

-C < -17

multiply both sides by -1

(-c) (-1) > (-17) (-1)

C > 17

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Mathematics Areas and perimeters

sweet-ann [11.9K]

Answer:

54.5 Sq. cm

Step-by-step explanation:

<em>area of a right angled triangle = 1/ 2 × base × height </em>

(here, the base is 8 and height is 6)

= 1/ 2 × 6 × 8

= 4 × 6

<u>= 24 Sq. cm</u>

<em>area of a circle</em> = $\pi$ r²

(r is the radius that is diameter/ 2, and the value of $\pi$ is 3. 14)

r = 10/ 2 = 5 cm

= 3. 14.× (5)²

= 3.14 × 25

<u>= 78.5 Sq. cm</u>

area of the shaded region = area of Circle - area of triangle

= 78.5 - 24

that is option 3

7 0

2 years ago

Will mark as brainliest (must show work)

Sever21 [200]

Answer:

Step-by-step explanation:

With those problems if you are not given a picture is good we draw one.

Because an angle bisector forms 2 congruent angles and because is given that < XVY ≅ < YVW then

m < XVY = m < YVW

2x+7 = x+15 , subtract x and 7 from both sides to isolate the like terms

2x-x = 15-7, combine like terms

x = 8

From the picture and the given we see that

m < XVW = m < XVY + m < YVW

m < XVW = 2x+7 + x+15 , combine like terms

m < XVW = 3x + 22, substitute x for 8

m < XVW = 3*8 + 22

m < XVW = 46

Check our work:

m < XVY = 2x+7 = 2*8 +7 = 16 +7 = 23

m < YVW = x+15 = 8 +15 = 23

m < XVW = m < XVY + m < YVW = 23+23 =46

3 0

2 years ago

In a small town in the Himalayas, the average snowfall each month is 6 inches. July is the month with the lowest average snowfal

OLga [1]

The function that models the above behavior is given as:

f(x) = ACos(B(x + C)) + 6 ..... .......i

where x = 1 (that is January)

D (Average Snowfall) = 6.

Hence, the amplitude (A) is given as:
A = 6-1

= 5

<h3>What is a mathematical function?</h3>

A mathematical function is a function comprising of different variables that represent different factors or events and the relationships between them.

<h3>
During what period is there more than 10 inches of snowfall?</h3><h3 />

Recall that in a year we have 12 months. Hence, P = 12

→ P = (2π) /.B → (2π)/B = 12 → B = π/6

Taking the values of A and B and plugging them in, we ahve:

F(x) = 5 cos (π/6(x + C)) + 6................ii

This means that in July, we'll have

x = 7 and f(x) = 1

Therefore,

1 = 5cos ((π/6) (7 + C) + 6 → 5 Cos ((π/6) (7 + C) → -5cos ((π/6) (7 + C)

= -1

→ ((π/6) (7 + C) = π

→ 7 + C = 6

→ C = -1

Taking equation ii, we have

F(x) = 5cos ((π6) (x -1)) + 6

F (x) > 10

Therefore

5cos ((π/6) (x-1) + 6 > 10

→ cos ((π/6) (x-1) + 6 > 4/5

→ (π/6) (x-1) > Cos⁻¹ (4/5)

(π/6) (x-1) < 0.6435 and

(π/6) (x-1) > 5.6397

→ x < 2.23 and x > 11.77

The interpretation is that upwards of ten inches of snow fall in January, February, November, and December.

Learn more about functions at;
brainly.com/question/25638609
#SPJ1

3 0

1 year ago

A minor-league hockey team had the following scores for the start of the season: 3, 6, 2, 1, 2, 3, 0, 4, 5, 1, 5, 4.

ale4655 [162]

If the hockey team scores 9 goals in their next game, it means that; the mean would be the most affected.

<h3>How to find mean, median and mode?</h3>

We are given the data set as;

3, 6, 2, 1, 2, 3, 0, 4, 5, 1, 5, 4.

Let us rearrange in ascending order to get;

0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6.

The mean = (0 + 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6)/12 = 3

Median = (3 + 3)/2 = 3

Mode = 1, 2, 3, 3, 5

Now, if they score 9 goals in their next game, it means that;

Mean = 45/13 = 3.46

Median = 3

Mode remains the same

Thus, mean will be the most affected.

Read more about Mean, Median and Mode at; brainly.com/question/14532771

#SPJ1

4 0

1 year ago

Distance between parallel lines y=3x+10 and y=3x-20

Alecsey [184]

1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).

2. Use formula $d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$ to find the distance from point $(x_0,y_0)$ to the line Ax+By+C=0.

The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:

$d=\dfrac{|3\cdot 0-10-20|}{\sqrt{3^2+(-1)^2}}=\dfrac{30}{\sqrt{10}}=3\sqrt{10}$ .

3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.

Answer: $d=3\sqrt{10}$ .

4 0

2 years ago