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just olya [345]
3 years ago
6

This is for algebra 2 and I got stuck on this anyone know this?

Mathematics
1 answer:
AnnZ [28]3 years ago
5 0
You just simplify the fraction (so, 8/6 which is 4/3)

So your comes 4(square root of 35)/ 3 (square root of 15)
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Find the values of p and q such that 4x^2+12x=4(x+p)^2-q
-Dominant- [34]

Answer:

p = 1.5 and q = 9

Step-by-step explanation:

Expand the right side then compare the coefficients of like terms on both sides, that is

4(x + p)² - q ← expand (x + p)² using FOIL

= 4(x² + 2px + p²) - q ← distribute parenthesis

= 4x² + 8px + 4p² - q

Comparing coefficients of like terms on both sides

8p = 12 ( coefficients of x- terms ) ← divide both sides by 8

p = 1.5

4p² - q = 0 ( constant terms ), that is

4(1.5)² - q = 0

9 - q = 0 ( subtract 9 from both sides )

- q = - 9 ( multiply both sides by - 1 )

q = 9

7 0
3 years ago
G (r) = 25 – 3r<br> g(4) =
tensa zangetsu [6.8K]

Answer:

=25 - 3 × 4

= 25-12= 13

...............

8 0
3 years ago
Read 2 more answers
Laura just finished a new book about a time-traveling magician. She read the same amount every day and completed the book in jus
svp [43]
So you have 85 pages and 10 days.
You would do 85/10 to get 8.5 pages each day. Hope that’s right !
5 0
2 years ago
A short-wave radio antenna is supported by two guy wires, 150 ft and 180 ft long. Each wire is attached to the top of the antenn
Dafna1 [17]

The distance between the anchor points of the two guy wires holding radio antenna is 181 feet (to the nearest foot)

<h3>How to determine the distance between the two anchor points of the guy wire</h3>

The problem will be solved using SOH CAH TOA

let the distance between the 150 ft long guy wire and the radio antenna be x

let the distance between the 180 ft long guy wire and the radio antenna be y

cos 65° = x / 150

x = cos 65° * 150

x = 63.39 ft

The height of the antenna

sin 65° = height of antenna / 150

height of antenna = sin 65 * 150

height of antenna = 135.95

using Pythagoras theorem

(length of guy wire)² = (height of the antenna)² + (anchor distance)²

(anchor distance)² = 180² - 135.95²

anchor distance = √(180² - 135.95²)

anchor distance = 117.97

The anchor points distance apart

= 63.39 + 117.97

= 181 (to the nearest foot)

Learn more on Pythagoras theorem here:

brainly.com/question/29241066

#SPJ1

4 0
1 year ago
Find the area pf a trapezoid with a base of 26.7 ft , a height of 15.4 and top of 9.9 ft
evablogger [386]
Hello!

To find the area of the trapezoid, you use the formula A=1/2h(a+b), where a and b represent the two base lengths.

Since we already know the two bases and the height, we can just plug them into the equation to find the area.

A=1/2·15.4(26.7+9.9)
A=1/2·15.4(36.6)
A=1/2·563.64
A=281.82

The area is 281.82 ft².

I hope this helps!
8 0
3 years ago
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