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Artyom0805 [142]
3 years ago
14

What is the range of this function (I NEED AN ANSWER RLLY FAST PLSSSS)

Mathematics
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

d

Step-by-step explanation:

the first oval's domain and the second's range. so you just choose the choice that includes all the values in the ranger.

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Yvette is considering taking out a loan with a principal of $16,200 from one of two banks. Bank F charges an interest rate of 5.
umka21 [38]
Hi there
First find the monthly payment of each offer to see which monthly payment is lower
The formula of the present value of annuity ordinary is
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
Pv present value
PMT monthly payment
R interest rate
K compounded monthly 12
N time
Solve the formula for PMT
PMT=pv÷[(1-(1+r/k)^(-kn))÷(r/k)]

Bank F
PMT=16,200÷((1−(1+0.057÷12)^(
−12×8))÷(0.057÷12))
=210.53

Bank G
PMT=16,200÷((1−(1+0.062÷12)^(
−12×7))÷(0.062÷12))
=238.21

From the above the monthly payment of bank f is lower than the bank g
And since the lifetime of bank g is lower than bank f the answer is
b. Yvette should choose Bank F’s loan if she cares more about lower monthly payments, and she should choose Bank G’s loan if she cares more about the lowest lifetime cost.

Good luck!


6 0
3 years ago
Is grade level qualitative or quantitative?
Tanya [424]
It would be quantitive I believe
8 0
3 years ago
Is y=x2-1 linear or nonlinear
Kazeer [188]

Answer:

y=x2-1 is Not Linear

6 0
2 years ago
Estelle drew two parallel lines PQ and R0S intersected by a transversal KL, as shown below: Which theorem could Estelle use to s
Sophie [7]
Angle KMQ and angle RNL are  on apposite sides of the transversal and between above and below (exterior to) the parallel lines, so they are alternate exterior angles.  Theorem - Alternate exterior angles formed by parallel lines and a transversal have the same measure.

8 0
2 years ago
Read 2 more answers
The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.
andrezito [222]

Answer:

Given the equation: 3x^2+10x+c =0

A quadratic equation is in the form: ax^2+bx+c = 0 where a, b ,c are the coefficient and a≠0 then the solution is given by :

x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a} ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}

Simplify:

x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}

Also, it is given that the difference of two roots of the given equation is 4\frac{2}{3} = \frac{14}{3}

i.e,

x_1 -x_2 = \frac{14}{3}

Here,

x_1 = \frac{-10 + \sqrt{100- 12c}}{6} ,     ......[2]

x_2= \frac{-10 - \sqrt{100- 12c}}{6}       .....[3]

then;

\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}

simplify:

\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}

or

\sqrt{100- 12c} = 14

Squaring both sides we get;

100-12c = 196

Subtract 100 from both sides, we get

100-12c -100= 196-100

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve x_1 , x_2

x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}

or

x_1 = \frac{-10 + \sqrt{100- 96}}{6} or

x_1 = \frac{-10 + \sqrt{4}}{6}

Simplify:

x_1 = \frac{-4}{3}

Now, to solve for x_2 ;

x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}

or

x_2 = \frac{-10 - \sqrt{100- 96}}{6} or

x_2 = \frac{-10 - \sqrt{4}}{6}

Simplify:

x_2 = -2

therefore, the solution for the given equation is: -\frac{4}{3} and -2.


3 0
2 years ago
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