All categories

2 years ago

I’ve been stuck on this for a few minutes. I’m not really sure what to do..

Mathematics

1 answer:

rewona [7]2 years ago

7 0

Answer:

Step-by-step explanation:

You know that the three angles of a triangle add up to 180°. If one angle is 38° and another angle is 67°, then the third angle must be 180°-38°-67° = 75°.

You have a straight line across the bottom, and three angles meeting in the middle of it. Those three angles add up to 180°. One of them is 75°, another is 50°, so the third angle must be 180°-75°-50° = 55°.

That leaves you with one unknown angle. ? = 180°-40°-55° = 85°

You might be interested in

Select ALL the numbers that are greater than 2.

ivanzaharov [21]

Answer: B and D

8/9 • 2 = 1.77

13/11 • 2 = 2.36

3.7 • 0.5 = 1.85

hope this helped :)

6 0

2 years ago

A park contains 210 trees on 40 acres. What is rate of trees per acre at the park?

love history [14]

Answer:

the answer would be 210:40

Step-by-step explanation:

5 0

2 years ago

Which of the following equations represents the line with a slope of -7/4 and a y-intercept of -11?

san4es73 [151]

Answer:

y = -7/4x - 11

Step-by-step explanation:

formula is y = mx + b

m is the slope

b is the y intercept

y = -7/4x -11

which is the first one

3 0

1 year ago

Justin Bieber was a gangster but ever since Justin Bieber got with Hayley he tried straight not straight as in lgtbq

masha68 [24]

TES YES WE I DONT KMOW HOW TO RESPOND TO THIS BUT YESS

8 0

2 years ago

Read 2 more answers

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$