MATHEMATICS HIGH SCHOOL

What is the measure of angle MBT

Answers

Answer 1

Answer:

Answer : The measure of angle MBT is 28°.

Step-by-step explanation :

As we see that the ∠NBM is equal to ∠MBT.

That means,

∠NBM = ∠MBT

Given:

∠NBM = (17x - 6)°

∠MBT = (5x + 18)°

According to question:

∠NBM = ∠MBT

(17x - 6)° = (5x + 18)°

17x - 5x = 18° + 6°

12x = 24°

x = 2°

So,

∠NBM = (17x - 6)° = [(17 × 2) - 6]° = 28°

∠MBT = (5x + 18)° = [(5 × 2) + 18]° = 28°

Hence, the measure of angle MBT is 28°.

Answer 2

Answer:

The measure of angle MBT is $\[ \text{Angle MBT} = 168 - 22x \].$

To find the measure of angle MBT, we can use the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180 degrees.

In triangle MBT, we have three angles: angle MBT (which we want to find), angle N (measured as (17x - 6)°), and angle BMT (measured as (5x + 18)°). According to the angle sum property, the sum of these angles is 180 degrees:

$\text{Angle MBT} + \text{Angle N} + \text{Angle BMT} = 180^\circ$

Substituting the given angle measures, we get:

$\text{Angle MBT} + (17x - 6)° + (5x + 18)° = 180°$

Now, combine like terms:

$\text{Angle MBT} + 22x + 12 = 180$

To solve for x, subtract 12 from both sides:

$\text{Angle MBT} + 22x = 168$

Finally, divide by 22 to solve for x:

$\text{Angle MBT} = 168 - 22x$

The value of x determines the measure of angle MBT. Without the specific value of x, we cannot determine the exact measure of angle MBT. If you provide the value of x, I can help you find the precise measure of angle MBT.

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Suppose an infection spreads according to the following data.What type of regression line is the best fit for this data?
A) linear
B) power
C) exponential
D) linear and exponential

Answers

Answer-

Exponential regression line is the best fit for the data.

Solution-

Taking

x = input variable,

y = output variable.

Taking the data from the table, regression models were generated using Excel.

As shown in the attachments, the co-efficient of determination (R²) is maximum for Exponential Regression model or more closer to 1.

As,

$0.904<0.957<0.974$

$R^2_{\text{Linear}}<R^2_{\text{Power}}<R^2_{\text{Exponential}}$

The more closer the value of R² to 1, the better the regression model and the best fit line is.

In general also, when we consider growth or decay, we follow the exponential function approach.

Therefore, the exponential regression models should be followed and so exponential regression line is the best fit for the data.

This is exponential and linear. Hope this helps!

Brainliest!!! Pls help

Answers

answer:

The equation of line s is given as y = 2. To find the coordinates of point C of AA'B'C' for Rs, we need to consider the intersection of line s with the x-axis.

When the equation of a line is in the form y = c, where c is a constant, it represents a horizontal line parallel to the x-axis. In this case, the line y = 2 is a horizontal line passing through the y-coordinate 2.

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"i was trying to leave the question but accidently pressed add answer lol"

Alli <3

What Is the next number? 12 21 23 32 34 43

Answers

12 21 23 32 34 43
so, we add 11 to 12 and get 23, then we add 11 to 23 and get 34, now we have to add 11 to 34 and get 45, that is the next number:
12 21 23 32 34 43 45 54

The next number of the number series 12 21 23 32 34 43 would be; 54

We are given the series as;

12 21 23 32 34 43

First, we have to add 11 to 12

11 + 12 = 23,

then we add 11 to 23

11 + 23 = 34,

Now we have to add 11 to 34

11 + 34 =45,

This is the next number:

12 21 23 32 34 43 45 54

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Find m2V.S
T
(9x – 19)°
U
111°
((7x + 3)
X
(5x + 8)
128°
W

Answers

Answer:

103

Step-by-step explanation:

6 sides

(6-2)180 =720

720 = S×T+U+V+W+X

S=90

720 = 90 + (9X-19) + (111) + (5X+8) + (128) + (7X+3)

COMBINE LIKE TERMS

720= (90-19+111+8+128+3) + (9X+5X+7X)

720 = 321 + 21X

720-321 = 321-321 + 21X

399 = 21X

19 = X

mV = 5X + 8

mV = 5(19) + 8

mV = 95 + 8

mV = 103

Convert the following fraction into its decimal equivalent: 3/8

Answers

3/8

Just divide to find the equal decimal.

3/8 = 0.375

Answer: 0.375 ✌️

the answer is 0.375. just divide

Consider the 1-to10^19 scale on which the disk of the Milky Way Galaxy fits on a football field. On this scale, how far is it from the sun to the alpa centauri (real distance:4.4 light-years) How big is the sun itself on this scale? Compare the sun's size on this scale to the actual size of a typical atom (about 10^-10 m in diameter).

Answers

Answer:

Sun is $0.4162708$ cm away from alpha century.

Sun is $1.391016*10^(-10)$ m.

Sun is 1.391016 time the size of an atom at this scale.

Step-by-step explanation:

Light year is a measure of distance. It is the distance light travels in an year.

Light year = $9.4607*10^(12)$ km

So 4.4 light years = $4.4*9.4607*10^(12)$ km

$41.62708*10^(12)$ km

Lets scale this down to the level of $1*10^(-19)$

$41.62708*10^(12)*1*10^(-19)$ km

= $41.62708*10^(-7)$ km

Change the units to centimeters:

$41.62708*10^(-7)*1*10^(5)$ cm

= $41.62708*10^(-2)$ cm

= $0.4162708$ cm

Therefore on the new scale sun is $0.4162708$ cm away from alpha century.

Diameter of the sun is 1.391016 million km

Lets change Sun's diameter to the new scale:

$1.391016*10^(6) *10^(-19)$ km

= $1.391016*10^(-13)$ km

Lets change kilometers in to meters:

$1.391016*10^(-13)*10^(3)$ m

= $1.391016*10^(-10)$ m

Therefore, sun is $1.391016*10^(-10)$ m

and an atom is $1*10^(-10)$

Therefore the sun is 1.391016 time the size of an atom at this scale.

Final answer:

On a 1-to-10^19 scale, the distance from the Sun to the Alpha Centauri is about 44 cm. On this same scale, the Sun itself would have a diameter of about 150 picometers, which is larger than a typical atom.

Explanation:

The 1-to-10^19 scale means for every actual meter in space, we represent 10^19 meters on our model. The Alpha Centauri is 4.4 light-years away from the sun. Considering 1 light-year equals to approximately 9.46x10^15 meters, the real distance from the sun to Alpha Centauri is about 4.16x10^16 meters. So, on the scale, this is about 0.44 meters or 44 cm.

The Sun's real size, with a diameter of 1.5 million kilometers or 1.5x10^9 meters, is represented as 1.5x10^-10 meters or 150 picometers on the scale. This is much bigger than an actual atom, which has a diameter of 0.1 to 0.5 nanometers or 100 to 500 picometers. Hence, on this scale, the Sun would be larger than a typical atom.