All categories

3 years ago

In the figure, m is parallel to n and m <4 = 125 degrees. Find the measures of the other angles.

Mathematics

1 answer:

Nesterboy [21]3 years ago

4 0

Answer:

m<1 = 55°

m<2 = 125°

m<3 = 55°

m<5 = 55°

m<6 = 125°

m<7 = 55°

m<8 = 125°

Step-by-step explanation:

m<4 = 125° (given)

✔️m<8 = m<4 (alternate exterior angles are congruent)

m<8 = 125° (substitution)

✔️m<1 = 180° - m<8 (supplementary angles/linear pair)

m<1 = 180° - 125° (substitution)

m<1 = 55°

✔️m<2 = m<8 (vertical angles are congruent)

m<2 = 125° (substitution)

✔️m<7 = m<1 (vertical angles are congruent)

m<7 = 55° (Substitution)

✔️m<3 = m<7 (alternate interior angles are congruent)

m<3 = 55° (substitution)

✔️m<5 = m<3 (vertical angles are congruent)

m<5 = 55°

✔️m<6 = m<4 (vertical angles)

m<6 = 125°

You might be interested in

What is this math problem??? g(x)=1/20x(x-100)

Fudgin [204]

Answer:

the x intercept is (0,100) and the y intercept is (0,0)

Step-by-step explanation:

g(x)=1/20x(x-100)

g(0)=1/20(0)(0-100)

g(0)=1/20*0(0-100)

g(0)=100

6 0

3 years ago

A movie theater has a seating capacity of 329. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.

kipiarov [429]

Answer:

170 children

74 students

85 adults

Step-by-step explanation:

Given

Let:

$C = Children; S = Students; A = Adults$

For the capacity, we have:

$C + S + A = 329$

For the tickets sold, we have:

$5C + 7S + 12A = 2388$

Half as many as adults are children implies that:

$A = \frac{C}{2}$

Required

Solve for A, C and S

The equations to solve are:

$C + S + A = 329$ -- (1)

$5C + 7S + 12A = 2388$ -- (2)

$A = \frac{C}{2}$ -- (3)

Make C the subject in (3)

$C = 2A$

Substitute $C = 2A$ in (1) and (2)

$C + S + A = 329$ -- (1)

$2A + S + A = 329$

$3A + S = 329$

Make S the subject

$S = 329 - 3A$

$5C + 7S + 12A = 2388$ -- (2)

$5*2A + 7S + 12A = 2388$

$10A + 7S + 12A = 2388$

$7S + 22A = 2388$

Substitute $S = 329 - 3A$

$7(329 - 3A) + 22A = 2388$

$2303 - 21A + 22A = 2388$

$2303 +A = 2388$

Solve for A

$A = 2388 - 2303$

$A = 85$

Recall that: $C = 2A$

$C = 2 * 85$

$C = 170$

Recall that: $S = 329 - 3A$

$S = 329 - 3 * 85$

$S = 329 - 255$

$S = 74$

Hence, the result is:

$C = 170$

$S = 74$

$A = 85$

8 0

3 years ago

The integral of (5x+8)/(x^2+3x+2) from 0 to 1

Gnom [1K]

Compute the definite integral:
integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx

Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
= integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx

Integrate the sum term by term and factor out constants:
= 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Evaluate the antiderivative at the limits and subtract.
(5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand 1/(x^2 + 3 x + 2), complete the square:
= (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx

For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds

Factor -1/4 from the denominator:
= (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds

Factor out constants:
= (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds

Factor -1 from the denominator:
= (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds

For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
= (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp

Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
= (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5

Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
= (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)

Which is equal to:

Answer: = log(18)

5 0

3 years ago

Write an equation in point slope form of the line that passes through the given point and the given slope m

victus00 [196]

Answer:

The answer is

$\huge y - 8 = 4(x + 9) \\$

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have the final answer as

$y - 8 = 4(x + 9)$

Hope this helps you

3 0

3 years ago

Line c has an equation of y = 4x + 9. Line dis parallel to line c and passes through (-4,-4).

natta225 [31]

Answer:

$\large\boxed{\sf y = 5x +16}$

Step-by-step explanation:

Here it is given that a line c has a equation of ,

$\sf\qquad\longrightarrow y = 4x + 9$

And there is another line d which is parallel to line c and passes through the point (-4,-4) . And we need to find out the equation of the line .

Firstly we know that the slope of two parallel lines is same . So on comparing the given line to the slope intercept form of the line which is y = mx + c , we have ;

$\sf\qquad\longrightarrow m = 4$