All categories

3 years ago

Geometry work help find the value of x

Mathematics

1 answer:

Oxana [17]3 years ago

8 0

Answer:

110⁰!!

Step-by-step explanation:

so this is an isosceles triangle! the base angles are always equal to each other. since 35 is a vertical angle, the base angle vertical to that would be 35. and since base angles ALWAYS equal each other, they would both be 35.

35+35=70, 70 is the base angles added up

180-70=110

c=110

You might be interested in

Assume the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons.

Rama09 [41]

The rate of change is 16.8 million tons per year

Solution:

Given that, the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons

The average rate of change is given by formula:

$\text{ Average rate of change } = \frac{\text{ change in value }}{\text{ number of years }}$

Value in 1950 = 12 million tons

Value in 1955 = 96 million tons

Change in value = value in 1955 - value in 1950

Change in value = 96 million - 12 million = 84 million tons

Number of years = 1950 to 1955 = 5 years

Substitute the given values in formula,

$\text{ Average rate of change } = \frac{84 \text{ million tons}}{5} = 16.8 \text{ million tons}$

Thus rate of change is 16.8 million tons per year

5 0

3 years ago

The length of a social media interaction is normally distributed with a mean of three minutes and a standard deviation of 0.4 mi

alexandr1967 [171]

It depends on Y and x cause school said so

3 0

3 years ago

V = the square root of 2as Solve for s

Gnom [1K]

As, v=√2as
Squaring both sides,
v²=2as
∴ s= v²/2a

6 0

3 years ago

(−2k 3 −7k 2 +5k)+(6k 2 +3k)=

shusha [124]

Answer: 5k +7

Step-by-step explanation: -2k + -7k + 5k is -4K and 2+3=5 -4K +5 and 6k +3k= 9k + 2 then u do -4K plus 9k and get 5k next, 5+2

7 0

3 years ago

Y=arccos(1/x) Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

3 years ago