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3 years ago

You put $8000 in an account. The account earns $2400 simple interest in 12 years. What is the annual interest rate

Mathematics

2 answers:

kvasek [131]3 years ago

8 0

Answer:

Step-by-step explanation:

bc i just did it

leva [86]3 years ago

7 0

Answer:

The interest rate was 2.5% per year.

Step-by-step explanation:

You might be interested in

There are 3 different roads from city a to city b and 2 different roads from city b to city

Nat2105 [25]

$3\cdot2=6$

7 0

2 years ago

A ball is thrown in the air from a platform that is 96 feet above ground level with an initial vertical velocity of 32 feet per

pishuonlain [190]

Answer:

y = -16 (x - 1)^2 + 112

The object lands on the ground in approximately 3.6s

Explanation:

The equation given is that of a parabola.

Now the maximum (local) point of a parabola is the vertex. Therefore, if we want to rewrite our function in the form that would be used to find the maximum height, then that form must be the vertex form of a parabola.

The vertex form of a parabola is

$y=a(t-h)^2+k$

where (h, k) is the vertex.

The only question is, what is the vertex for our function h(t)?

Remember that if we have an equation of the form

$y=ax^2+bx+c$

then the x-coordinate of the vertex is

$h=-\frac{b}{2a}$

Now in our case b = 32 and a = -16; therefore,

$h=\frac{-32}{2(16)}=1$

We've found the value of the x-coordinate of the vertex. What about the y-coordinate? To get the y-coordinate, we put x = 1 into h(t) and get

$k=-16(1)+32(1)+96=112$

Hence, the y-coordindate is k = 112.

Therefore, the vertex of the parabola is (1, 112).

With the coordinates of the vertex in hand, we now write the equation of the parabola in vertex form.

$h(t)=a(t-1)^2+112$

The only problem is that we don't know what the value of a is. How do we find a?

Note that the point (0, 96) lies on the parabola. In other words,

$h(0)=-16(0)^2+32(0)+96=96$

Therefore, the vertex form of the parabola must also contain the point (0, 96).

Putting in t = 0, h = 96 into the vertex form gives

$96=a(0-1)^2+112$ $96=a+112$

subtracting 112 from both sides gives

$a=-16$

With the value of a in hand, we can finally write the equation of the parabola on vertex form.

$\boxed{h\mleft(t\mright)=-16\left(t-1\right)^2+112.}$

Now when does the object hit the ground? In other words, for what value of t is h(t) = 0? To find out we just have to solve the following for t.

$h(t)=0.$

We could either use h(t) = -16t^2 + 32t + 96 or the h(t) = -16(t - 1)^2 + 112 for the above equation. But it turns out, the vertex form is more convenient.

Thus we solve,

$-16\left(t-1\right)^2+112=0$

Now subtracting 112 from both sides gives

$-16(t-1)^2=-112$

Dividing both sides by -16 gives

$(t-1)^2=\frac{-112}{-16}$ $(t-1)^2=7$

taking the square root of both sides gives

$t-1=\pm\sqrt{7}$

adding 1 to both sides gives

$t=\pm\sqrt{7}+1$

Hence, the two solutions we get are

$t=\sqrt{7}+1=3.6$ $t=-\sqrt{7}+1=-1.6$

Now since time cannot take a negative value, we discard the second solution and say that t = 3.6 is our valid solution.

Therefore, it takes about 3.6 seconds for the object to hit the ground.

3 0

10 months ago

Please help me solve this I will give u brainlst :)

NARA [144]

Equation: 2(.25) + p(.75) = 2.75
Answer: 3 pencils

7 0

2 years ago

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$

$= - \dfrac{1}{2} cos (4) + \dfrac{1}{2} (1) \\ \\ = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\ = \mathbf{0.82682}$

5 0

2 years ago

If f(x) is a third degree polynomial function and g(x) = f(–2x), which of the following is NOT true?

valina [46]

Answer:

f(x) and g(x) have the same x-intercepts (is <em>not true</em>)

Step-by-step explanation:

g(x) is a reflection across the y-axis and a horizontal compression of f(x). In general those transformations will move the x-intercepts. (The y-intercept and the number of x-intercepts will remain unchanged.)

_____

<em>Comment on the question/answer</em>

f(x) = x^3 is a 3rd degree polynomial. When transformed to g(x) = -8x^2, its only x-intercept (x=0) remains the same. The answer above will not apply in any instance where the only x-intercept is on the line of reflection. (The question is flawed in that it does not make any exception for such functions.)

8 0

3 years ago