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Tanzania [10]
2 years ago
5

Please answer the question attached, marking brainliest (serious answers only)

Mathematics
2 answers:
Mandarinka [93]2 years ago
8 0

Step-by-step explanation:

By Exterior Angle Theorem,

81° + b = 117°.

=> b = 117° - 81° = 36°.

Elden [556K]2 years ago
5 0

Answer:

b = 36

Step-by-step explanation:

a+117 = 180 ( being straight angle)

a = 180 - 117

a = 63

Now,

81 + 63 + b = 180 ( being sum of interior of triangle)

144 + b= 180

b = 180 - 144

b 36

You might be interested in
Divide 7/15 by 3/5. <br> A. 75/21<br> B. 7/9<br> C. 7/25<br> D. 21/75
Digiron [165]
In this question there is nothing complicated. Only thing is to know the way fractions can be divided. Once that is known the problem would be one of the easiest to solve. Now let us get back to the problem and look at all the information's that are given in the question.
Divide 7/15 by 3/5 = (7/15)/(3/5)
                              = (7 * 5)/(15 * 3)
                               = (35/45)
Dividing the numerator and the denominator by 5 for simplifying purpose, we get
                               = 7/9
So from the above deduction we can easily conclude that 7/9 is the correct answer and option "B" is the correct option among all the options given in the question.
5 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST!
siniylev [52]

Answer:

48x² + 24x + 3

Step-by-step explanation:

Width = (4x + 1) in

Length = 3 *width = 3*(4x + 1) = 3*4x + 3*1

           = (12x + 3) in

Area of rectangle = length * width

                             = (12x + 3) (4x + 1)

                             = 12x *4x  + 12x *1  + 3*4x  + 3*1

                             = 48x² + 12x + 12x + 3

                              = 48x² + 24x + 3

8 0
3 years ago
Which graph shows a dilation of the triangle with a scale factor of 3
8090 [49]

I'd say the 3rd one. I haven't done this stuff in 2 years ... me forgot. But if it's not right, then the 1st one!

I want you to know you're smart and you can do this! Good luck!

3 0
3 years ago
Read 2 more answers
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 4 (a) Find the interval on which
kozerog [31]

Answer:

a) The function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Written in interval form

(-∞, -1.45) and (3.45, ∞)

- The function, f(x) is decreasing at the interval (-1.45 < x < 3.45)

(-1.45, 3.45)

b) Local minimum value of f(x) = -78.1, occurring at x = 3.45

Local maximum value of f(x) = 10.1, occurring at x = -1.45

c) Inflection point = (x, y) = (1, -16)

Interval where the function is concave up

= (x > 1), written in interval form, (1, ∞)

Interval where the function is concave down

= (x < 1), written in interval form, (-∞, 1)

Step-by-step explanation:

f(x) = x³ - 6x² - 15x + 4

a) Find the interval on which f is increasing.

A function is said to be increasing in any interval where f'(x) > 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

the function is increasing at the points where

f'(x) = 3x² - 6x - 15 > 0

x² - 2x - 5 > 0

(x - 3.45)(x + 1.45) > 0

we then do the inequality check to see which intervals where f'(x) is greater than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

So, the function (x - 3.45)(x + 1.45) is positive (+ve) at the intervals (x < -1.45) and (x > 3.45).

Hence, the function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Find the interval on which f is decreasing.

At the interval where f(x) is decreasing, f'(x) < 0

from above,

f'(x) = 3x² - 6x - 15

the function is decreasing at the points where

f'(x) = 3x² - 6x - 15 < 0

x² - 2x - 5 < 0

(x - 3.45)(x + 1.45) < 0

With the similar inequality check for where f'(x) is less than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

Hence, the function, f(x) is decreasing at the intervals (-1.45 < x < 3.45)

b) Find the local minimum and maximum values of f.

For the local maximum and minimum points,

f'(x) = 0

but f"(x) < 0 for a local maximum

And f"(x) > 0 for a local minimum

From (a) above

f'(x) = 3x² - 6x - 15

f'(x) = 3x² - 6x - 15 = 0

(x - 3.45)(x + 1.45) = 0

x = 3.45 or x = -1.45

To now investigate the points that corresponds to a minimum and a maximum point, we need f"(x)

f"(x) = 6x - 6

At x = -1.45,

f"(x) = (6×-1.45) - 6 = -14.7 < 0

Hence, x = -1.45 corresponds to a maximum point

At x = 3.45

f"(x) = (6×3.45) - 6 = 14.7 > 0

Hence, x = 3.45 corresponds to a minimum point.

So, at minimum point, x = 3.45

f(x) = x³ - 6x² - 15x + 4

f(3.45) = 3.45³ - 6(3.45²) - 15(3.45) + 4

= -78.101375 = -78.1

At maximum point, x = -1.45

f(x) = x³ - 6x² - 15x + 4

f(-1.45) = (-1.45)³ - 6(-1.45)² - 15(-1.45) + 4

= 10.086375 = 10.1

c) Find the inflection point.

The inflection point is the point where the curve changes from concave up to concave down and vice versa.

This occurs at the point f"(x) = 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

f"(x) = 6x - 6

At inflection point, f"(x) = 0

f"(x) = 6x - 6 = 0

6x = 6

x = 1

At this point where x = 1, f(x) will be

f(x) = x³ - 6x² - 15x + 4

f(1) = 1³ - 6(1²) - 15(1) + 4 = -16

Hence, the inflection point is at (x, y) = (1, -16)

- Find the interval on which f is concave up.

The curve is said to be concave up when on a given interval, the graph of the function always lies above its tangent lines on that interval. In other words, if you draw a tangent line at any given point, then the graph seems to curve upwards, away from the line.

At the interval where the curve is concave up, f"(x) > 0

f"(x) = 6x - 6 > 0

6x > 6

x > 1

- Find the interval on which f is concave down.

A curve/function is said to be concave down on an interval if, on that interval, the graph of the function always lies below its tangent lines on that interval. That is the graph seems to curve downwards, away from its tangent line at any given point.

At the interval where the curve is concave down, f"(x) < 0

f"(x) = 6x - 6 < 0

6x < 6

x < 1

Hope this Helps!!!

5 0
3 years ago
Select the point that is a solution to the system if inequalities y&lt; x^2+3 y&gt;x^2-4?
stira [4]

Answer:

 

Step-by-step explanation:

you can  potting a picture to have a look

3 0
3 years ago
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