All categories

3 years ago

How do you do this I don’t get it

Mathematics

1 answer:

Ierofanga [76]3 years ago

6 0

Answer:

x= 65 z= 38 y= 77

Step-by-step explanation:

x=65° (opposite angles are equal)

z= 180°-[71+71 (bc the other side is also equal to 71)]

z= 180°-142°=38°

y= 180 - [sum of angle x (65°) and angle z (38°)]

y= 180 - 103= 77°

to make sure ur answer is correct the sum of angle x, y and z MUST = 180

lets check:

65° + 38° + 77° = 180°

You might be interested in

3/4x+3y=12 for x when y=5

Karolina [17]

Solve for x by simplifying both sides of the equation, then isolating the variable.
x = -4
Hope this helps! :)
Happy New Years!

4 0

3 years ago

Which expression is equivalent to 5^4(5^-^3)^2?<br> 5^1 <br> 5^2<br> 1/5^1<br> 1/5^2

inessss [21]

Answer:

1 5th ^2.

Step-by-step explanation:

8 0

3 years ago

Marking Brainliest for first right answer

BabaBlast [244]

The answer is Between 12 and 26

3 0

3 years ago

Read 2 more answers

Vanessa tried to prove that triangle KLM is congruent to triangle MNK. What is the first error Vanessa made in her proof?

Natali [406]

Answer:

Step-by-step explanation:

Vanessa only established some of the necessary conditions

5 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

4 years ago