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Pani-rosa [81]
1 year ago
11

Triangle PQR has coordinates P (2, 4), Q (-2, 4), R (0,-6).

Mathematics
2 answers:
jok3333 [9.3K]1 year ago
7 0

Answer:dec I think

Step-by-step explanation:

alexdok [17]1 year ago
4 0

Answer: P’ (5,10) , Q’ (-5,10) , R’ ( 0,-15)

Step-by-step explanation:

Just multiply 2.5 by the x,y

You might be interested in
A bullet is shot straight upward with an initial speed of 925 ft/s. The height of the bullet after t seconds is modeled by the f
never [62]

Answer:

462 ft/s  

Step-by-step explanation:

The height of the bullet is modeled by:

h(t) = -16t² + 925t------------------------------------ (1)

The speed of the bullet is modeled by the first derivative of equation (1)

dh/dt= -32t + 925------------------------------------ (2)

At maximum height,  dh/dt = 0

0 = -32t  + 925

32t = 925

  t  =   925/32

     =   28.91 seconds

This means that the total time to reach maximum height is 28.91 seconds

Substituting into equation (1) we can calculate the maximum  height:

h = -16 (28.91)² +925 (28.91)

  =   -13,372.61 + 26,742.675

  =  13,367.065 ‬ft

Average speed =  Total distance/ Total time

                         =    13,367.065/28.91

                         =  462.368

                         ≈ 462 ft/s  

7 0
2 years ago
Which function has a Vertex at the origin f(x)= (x+4)^2, f(x) x(x-4), f(x)=(x-4)(x+4), f(x)=-x^2
Studentka2010 [4]
First of all, we need to know what is the vertex means which is the maximum or minimum point of a parabola and the formula will be:
x=-b/2a
Where b and a from
f(x)=ax^2+bx+c
So do find which function has a vertex of origin. Let's find the vertex of all the function that we had:
f(x)=(x+4)^2
f(x)=(x+4)(x+4)
f(x)=x^2+8x+16
x=-b/2a
x=-8/2(1)
x=-8/2
x=-4
Not the right answer because the vertex needs to be origin which is x=0

f(x)=x(x-4)
f(x)=x^2-4x
x=-b/2a
x=-(-4)/2(2)
x=4/4
x=1
Not the right answer

f(x)=(x-4)(x+4)
f(x)=x^2-16
x=-0/2(1)
x=0
Yay! This is the right answer. As a result, f(x)=(x-4)(x+4) is your final answer. Hope it help!
7 0
1 year ago
Your company has just taken out a 1-year installment loan for $72,500 at a nominal rate of 20.0% but with equal end-of-month pay
Mariulka [41]

Answer:

100% of the 2nd monthly payment go toward the repayment of principal.

Step-by-step explanation:

The loan taken is the Principal which is mentioned as $72,500 with interest at a nominal rate of 20%. Firstly, it is important to understand that nominal rate means <em>non-compounding </em>rate. Simply put will be a "<em>one-time charged" </em>rate on the loan. Since this is given as 20% of the Principal. It is calculated thus: \frac{20}{100} × \frac{72,500}{1}= $14,500. So the interest on the loan is $14,500. Added to the Principal the total amount to be paid back by the company becomes: $72,500 + $14,500 = $87,000. To pay back this amount at equal end-of-month installments in 1 year (12 months), we divide the total amount by 12. i.e \frac{87000}{12} = $7250. This means, the monthly payment will be $7,250. Since the monthly payment pays only 10% of the initial principal $72,500. By the second month only 20% of the Principal would have been paid. So all of the monthly payment will go towards repaying the principal

3 0
2 years ago
Let e1= 1 0 and e2= 0 1 ​, y1= 4 5 ​, and y2= −2 7 ​, and let​ T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps
Furkat [3]

Answer:

The image of \left[\begin{array}{c}4&-4\end{array}\right] through T is \left[\begin{array}{c}24&-8\end{array}\right]

Step-by-step explanation:

We know that T: IR^{2}  → IR^{2} is a linear transformation that maps e_{1} into y_{1} ⇒

T(e_{1})=y_{1}

And also maps e_{2} into y_{2}  ⇒

T(e_{2})=y_{2}

We need to find the image of the vector \left[\begin{array}{c}4&-4\end{array}\right]

We know that exists a matrix A from IR^{2x2} (because of how T was defined) such that :

T(x)=Ax for all x ∈ IR^{2}

We can find the matrix A by applying T to a base of the domain (IR^{2}).

Notice that we have that data :

B_{IR^{2}}= {e_{1},e_{2}}

Being B_{IR^{2}} the cannonic base of IR^{2}

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]   (Data of the problem)

T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]   (Data of the problem)

Writing the matrix A :

A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]

Now with the matrix A we can find the image of \left[\begin{array}{c}4&-4\\\end{array}\right] such as :

T(x)=Ax ⇒

T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]

We found out that the image of \left[\begin{array}{c}4&-4\end{array}\right] through T is the vector \left[\begin{array}{c}24&-8\end{array}\right]

3 0
2 years ago
Merle Fonda opened a new savings account. She deposited $40,000 at 10% compounded semiannually. At the start of the fourth year,
IrinaVladis [17]
Use compound interest formula  F=P(1+i)^n twice, one for each deposit and sum the two results.

For the P=$40,000 deposit,
i=10%/2=5%  (semi-annual)
number of periods (6 months), n = 6*2 = 12
Future value (at end of year 6),
F = P(1+i)^n = 40,000(1+0.05)^12 = $71834.253

For the P=20000, deposited at the START of the fourth year, which is the same as the end of the third year.
i=5% (semi-annual
n=2*(6-3), n = 6 
Future value (at end of year 6)
F=P(1+i)^n = 20000(1+0.05)^6 = 26801.913

Total amount after 6 years
= 71834.253 + 26801.913
=98636.17   (to the nearest cent.)
8 0
1 year ago
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