Answer:
<h2>10 weeks </h2>
Step-by-step explanation:
Step one:
given data
Ryan
number of baseball cards=150
number collected per week= 10
let the number of weeks be x
and the total be y
y=10x+150-----------------1
Sarah
number of baseball cards=200
number collected per week= 5
let the number of weeks be x
and the total be y
y=5x+200------------2
Step two:
Required
the number of weeks where both total will be the same
10x+150=5x+200
10x-5x=200-150
5x=50
divide both sides by 5
x=50/5
x=10 weeks
Answer:
<h2>Area = 14m²</h2>
<u>Step-by-step explanation:</u>
area of rectangle = length × breadth
area of rectangle = 4 × 2
area of rectangle = 8m²
area of Triangle = 1/2 × base × height
area of Triangle = 1/2 × 4 × 3
area of Triangle = 6m²
Total area = 8 + 6
Total area = 14m²
Answer:

Step-by-step explanation:
The domain of a function is all of the values that
can be under that specific function. In this case, we're asking what values of
allow
to exist.
In order for square roots to exist, the quantity under the square root must be greater than or equal to
, because you can't take the square root of a negative number. Therefore, we can write the following inequality to solve for
:

Solving this inequality, we get:

(Subtract
from both sides of the inequality to isolate
)
(Multiply both sides of the inequality by
to get rid of
's coefficient)
Hope this helps!
Ther eqaul becayse there are four quarts in a gallon
<h3>Given:-</h3>

<h3>To Find:-</h3>
<h3>Solution:-</h3>










we can also write it as ;

★<u> </u><u>Henceforth, the value of a and b are</u> :
→ a = 0
→ b = 1