Answer:



Step-by-step explanation:
-13+13r
-22+6r-7
bring all r and number one oppsite sides
13r - 6r
-22-7+13
7r
-16
r

mark me as branliest. ty
<span>2.8 x 10^6 N
We have 9.8 x 10^7 J or 9.8 x 10^7 kg*m^2/s^2 of work done through a distance of 35 meters. Since work is defined as force over distance, that means that:
F * 35m = 9.8 x 10^7 kg*m^2/s^2
Solving for F, gives
F * 35m = 9.8 x 10^7 kg*m^2/s^2
F = 9.8 x 10^7 kg*m^2/s^2 / 35m
F = 2.8 x 10^6 kg*m/s^2
F = 2.8 x 10^6 N
So the force exerted by the tugboat is 2.8x10^6 Newtons.</span>
Answer:
x = - 2, x = 12
Step-by-step explanation:
Given
x² - 10 x = 24 ( subtract 24 from both sides )
x² - 10x - 24 = 0 ← in standard form
To factorise the quadratic
Consider the factors of the constant term (- 24) which sum to give the coefficient of the x- term (- 10)
The factors are - 12 and + 2, since
- 12 × 2 = - 24 and - 12 + 2 = - 10, thus
(x - 12)(x + 2) = 0
Equate each factor to zero and solve for x
x - 12 = 0 ⇒ x = 12
x + 2 = 0 ⇒ x = - 2
First pic
a. ΔMNK ≅ ΔRTP
b. TR ≅ NM
c. x = 7
3x - 1 = 20
3x = 20 + 1
3x = 21
x = 21 ÷ 3
x = 7
Second pic
SAS (Side-Angle-Side congruence)
SSS (Side-Side-Side)
Hope this helps
1. 5 in and 1/3 in: Area = 5/3 in^2
2. 5 in and 4/3 in: Area= 20/3 in^2
3. 5/2 in and 4/3 in: Area=10/3 in^2
4. 7/6 in and 6/7 in: Area = 1 in^2
Step-by-step explanation:
<u>1. 5 in and 1/3 in</u>
Here,

<u>2. 5 in and 4/3 in</u>
Here,

<u>3. 5/2 in and 4/3 in</u>

<u>4. 7/6 in and 6/7 in</u>
<u>Let</u>

<u>Hence,</u>
1. 5 in and 1/3 in: Area = 5/3 in^2
2. 5 in and 4/3 in: Area= 20/3 in^2
3. 5/2 in and 4/3 in: Area=10/3 in^2
4. 7/6 in and 6/7 in: Area = 1 in^2
Keywords: Rectangle, Area
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