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likoan [24]
4 years ago
14

A company's stock was selling at

Mathematics
2 answers:
damaskus [11]4 years ago
8 0
The percent increase is 30%
Vlad [161]4 years ago
4 0
Hope it helps even tho someone already answered this!

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alexira [117]
It's A
because it can be into a decimal and then to an equation
8 0
3 years ago
Read 2 more answers
a highway is to be built between two towns, one of which lies 48.6 km south and 60.3 km west of the othe
Katena32 [7]

Answer:

whats the rest of the question or what's the available answers

8 0
1 year ago
Help me I need help please anyone
jarptica [38.1K]

Answer:

1st,4th,5th option.

Step-by-step explanation:

Let evaluate each option.

A segment bisector is a line segment, ray, and/or line that bisects a line into two congruent parts. LM splits JK and KH into congruent parts. The first option is correct.

A perpendicular bisector is a line segment,ray and/or line that intersects a line segment,ray at a right angle. We don't have a perpendicular angle here and so that isn't a option.

M isn't a vertex of congruent angles as there is none in this figure.

The fourth and fifth option are correct. They both are on the segment bisector so they split the figure segments into two congruent parts. Since they are on the line segment and bisects it, they are considered the midpoint or middle point of the figure side.

7 0
3 years ago
Conner works 8 1/2 hours and earns $80.75. If Zoey makes an hourly rate that is proportional and earns $118.75, how many hours d
Reptile [31]

Answer:

Step-by-step explanation:

In 8 hours 30m Corner earns $80.75

In how many hours Zoey earns $118.75

Hours Zoey works= 510*118.75/80.75(converted 8hours into minutes by multiplying with 60)

Solving we get 12 hours 30m.

8 0
4 years ago
Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal
babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

Now, the midpoints of these subintervals are:

\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11

Hence:

M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698

(b)

\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))

n=4, so :

Each subinterval has length :

\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557

8 0
3 years ago
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