Xz is 15 and the perimeter of triangle xwz is 50 what is the length of wz

Answers

Answer 1

Answer: Well if you have a congruent lines, and you have P = 50 and xz = 15 P = xz + 2 * wz so to find wz you need to subtract xz and divide by 2 50 - 15 = 35 and then divide by 2 to get 17.5

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Cam rode her bike 5 times as far as Dante did. Dante rode 187 meters farther than Michael did. Cam rode 15.25 Kilometers. How many meters did Michael ride? Explain how you know you got your answer

Answers

Let Dante's distant be x, Cam's will be 5x
Given that Cam rode 15.25 km
then
5x=15.25
x=3.05 km

thus Dante rod 3.05 km
but
1 km=1000m
thus
3.05 km=3050 m
But Dante rod 187 m more than Michael, the distance Michael will ride then will be:
3050-187
=2863 m

Can anybody help by forming and solving this equation (plz Show steps and worhing)A train has 180 passengers At the fist stop X get off and 24 get on At the second stop half get off and 40 get on There are now 134 passengers How many passengers got off at the 1st stop (X)

thks 4 any help

Answers

so a the start 180 passengers are on the train
at the first stop x get off and 24 get on or
180-x+24=204-x

at the second stop half get off and 40 get on or
204-x-([204-x]/2)+40 or
204-x-102+x/2+40
104-1/2x+40
144-1/2x
there are now 134 passengers
144-1/2x=134
subtract 134 from both sides
10-1/2x=0
add 1/2x to both sides
10=1/2x
mutliply both sides by 2/1 to make 1/2x=1/1x
20=x
20 got off at the first stop

Form an equation using information given
180 - x - 24 + 40 = 134
then simplify and solve
196 - x = 134
196 - 134 = x
62 = x
therefore 62 people got off at the first stop

Give the equation of the horizontal asymptote shown below.g(x)=7x^2/x^2+5
A. y= 7/5
B. y=7
C. y=5
D. x=7

Answers

Polynomial division yields

$(7x^2)/(x^2+5)=7-(35)/(x^2+5)$

As $x\to\pm\infty$ , you have the remainder term approaching $0$ , so the horizontal asymptote is the line $y=7$ .

Answer:

the answer on a*p*e*x is y=7

Simplify 1/4(8m - 4n) + 1/3(6m + 3n)

Answers

Answer:

The simplified of $(1)/(4)(8m-4n)+ (1)/(3)(6m+3n)$ is 4m .

Step-by-step explanation:

As given

$= (1)/(4)(8m-4n)+ (1)/(3)(6m+3n)$

L.C.M of (4,3) = 12

$= (3* (8m-4n)+4* (6m+3n))/(12)$

$= (3* 8m-3* 4n+4* 6m+4* 3n))/(12)$

$= (24m-12n+24m+12n)/(12)$

$= (24m+24m)/(12)$

$= (48m)/(12)$

= 4m

Therefore the simplified of $(1)/(4)(8m-4n)+ (1)/(3)(6m+3n)$ is 4m .

1/4(8m - 4n) + 1/3(6m + 3n)=

Distribute the fractions through the parentheses

2m - n + 2m + n =

Combine like terms

2m + 2m - n + n = 4m

Simplified Answer is 4m

If a company has 150 shares of common stock and $15,000.00 to be distributed to its holders, how much would each share receive?

Answers

If there are a total of 150 shares of common stock that total $15,000.00 total, you will need to divide the total price by the number of shares to determine the price per share. In this example, it would be as follows: $15,000.00/150 shares = $100/share. Each share would receive $100.

Which statements are true for the functions g(x) = x2 and h(x) = –x2 ? Check all that apply.For any value of x, g(x) will always be greater than h(x).
For any value of x, h(x) will always be greater than g(x).
g(x) > h(x) for x = -1.
g(x) < h(x) for x = 3.
For positive values of x, g(x) > h(x).
For negative values of x, g(x) > h(x).

Answers

if x=0 then they have same value

1st and 2nd options are out

for x=-1
g(-1)=1
h(-1)=-1
3rd is true

4th
false

for all values except zero, g(x)>h(x)

correct ones are

g(x) > h(x) for x = -1.
For positive values of x, g(x) > h(x).
For negative values of x, g(x) > h(x).

Answer: g(x) > h(x) for x = -1.

For positive values of x, g(x) > h(x).

For negative values of x, g(x) > h(x).

Step-by-step explanation:

Given functions: $g(x)=x^2$ and $h(x)=-x^2$

When x=0, $g(0)=0^2=0$ and $h(0)=-0^2=0$

∴ at x=0, g(x)=h(0)

Therefore the statements "For any value of x, g(x) will always be greater than h(x)." and "For any value of x, h(x) will always be greater than g(x)." are not true.

When x=-1, $g(-1)=(-1)^2=1$ and $h(-1)=-(-1)^2=-1$

∴g(x) > h(x) for x = -1. ......................(1)

When x=3, $g(3)=(3)^2=9$ and $h(3)=-(3)^2=--9$

∴ g(x) > h(x) for x = 3....................(2)

⇒g(x) < h(x) for x = 3. is not true.

From (1) and (2),

For positive values of x, g(x) > h(x).

For negative values of x, g(x) > h(x).

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