Answer:
V=14
Step-by-step explanation:
Well its a supplementary angle if you add everything together.
We know theres a 90 degrees angle and a 34 degree angle.
90+34=124
124+4v=180
Since if you add everything up its a straight line
Solve the equation normally
56=4v
14=v
Answer:
A
Step-by-step explanation:
In the histogram described by the first option, the peak (18) is located at 40-50. Also, a frequency of zero is observed for range 10-20. The cluster from 20–50 means that most of the data is between this range, in fact, only 1 point is located at range 0 - 10, outside this cluster.
Answer: $ 4,515
Step-by-step explanation:
The cost of the car she wanted to buy = $35,550
Her savings = $20,700
That means she needs( $35,550 - $20,700) more in order for her to buy the car.
Amount borrowed from the finance company = $35,550 - $20,700
= $ 14,850
Since she now owes the financial company a total of $19,365, then the interest accrued by the loan she borrowed is given by:
$19,365 - $ 14,850
= $4,515
Answer:
Width = 7 cm
Length = 15 cm
Step-by-step explanation:
Let width = w
Length = w + 8
Area of rectangle = 105 sq. cm
length * breadth = 105
(w + 8)*w = 105
Use distributive property
w² + 8w = 105
w² + 8w - 105 = 0
Sum = 8
Product = -105
Factors = 15 , -7 {15 +(-7) = 8 & 15*(-7) = -105}
w² - 7m + 15m - 105 = 0
w(w - 7) + 15(m - 7) = 0
(w - 7)(w + 15) = 0
Ignore w + 15= 0 as measurements will not be in negative value
w - 7 = 0
w = 7
Width = 7 cm
Length = 7 + 8 = 15 cm
<h2>T
he car is about 6.6 years old.</h2>
Step-by-step explanation:
Given : An equation for the depreciation of a car is given by , where y = current value of the car, A = original cost, r = rate of depreciation, and t = time, in years. The value of a car is half what it originally cost. The rate of depreciation is 10%.
To find : Approximately how old is the car?
Solution :
The value of a car is half what it originally cost i.e.
The rate of depreciation is 10% i.e. r=10%=0.1
Substitute in the equation,
Taking log both side,
Therefore, the car is about 6.6 years old.