All categories

3 years ago

Can someone tell me how to solve this?

Mathematics

1 answer:

liberstina [14]3 years ago

4 0

These two angles are equivalent because they're corresponding angles created by two parallel lines.

Since they're equivalent, we can set them equal to each other.

23x - 5 = 21x + 5

Now solve for x

23x - 5 = 21x + 5

Add 5 to both sides

23x = 21x + 10

Subtract 21x from both sides

2x = 10

Divide both sides by 2

x = 5

Now, plug in 5 as "x" in one of the equations

Angle 1 = (23*5) - 5

= 115 - 5

= 110 degrees

Since they're equivalent, Angle 2 should also be 110 degrees, but let's check in case.

Angle 2 = (21*5) + 5

= 105 + 5

= 110 degrees

You might be interested in

Can someone help me please it is from numbers 16 to 19 please someone

TiliK225 [7]

The correct answers are:

16) Mean = 10.25; Median = 10; Range = 17
17) The range.
18) 10
19) 5+5+6+6+7+8+8+8+9+9+10+10+10+10+10+11+12+12+12+13+13+15+15+22 = 246.

Explanation:

16) To find the mean, we first find the sum of the data values in the set:
5+5+6+6+7+8+8+8+9+9+10+10+10+10+10+11+12+12+12+13+13+15+15+22 = 246.

Next we divide this sum by the number of data points, 24:
246/24 = 10.25.

The median is the middle point of a data set. Since there are 24 points, this will be between two values. These two values are 10 and 10; the median is 10.

The range is found by subtracting the highest and lowest values:
22-5 = 17.

17) The mean is changed; the sum of the data values is 246. Taking 22 out of this, the sum is now 223, and we have 23 data points instead of 24; 223/23 = 9.7 for the new mean.
The median does not change; it is still 10.
The range changes; the highest value is now 15, and the lowest is 5: 15-5=10. The range is changed the most.

18) The mean is affected by the outlier and the median is not, so we use the median. This means the answer is 10.

19) The expression for the total number of cars sold is found by adding together all of the points.

6 0

3 years ago

2/5 of a class are girls. 3/4 of the boys are called Paul. What fraction of the whole class is not called Paul?

Novay_Z [31]

Answer:

1/4

Step-by-step explanation:

3/4 + 1/4 = 4/4

7 0

3 years ago

Read 2 more answers

12x⁴y²divided by 3x³y to the 5 th power....Please help me out....

ki77a [65]

To divide the expression we proceed as follows:
12x⁴y²÷3x³y⁵
=12x⁴y²/3x³y⁵
=(12÷3)×(x⁴÷x³)×(y²÷y⁵)
when you divide number with the same base, you subtract the numerators:
=4×(x⁴⁻³)×(y²⁻⁵)
simplifying this we get
=4xy⁻³
Answer: 4xy⁻³

5 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago

A globe in the shape of a sphere has a radius of 3/4 ft. What is the volume of the globe?

mafiozo [28]

Answer:

1.768

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers