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anzhelika [568]
3 years ago
7

The sum of two numbers is 90. The larger

Mathematics
1 answer:
Jobisdone [24]3 years ago
8 0

Answer:

Small number = 19

Larger no. = 3(19)+14 = 71

Step-by-step explanation:

x + 3x + 14 = 90

4x + 14 = 90

4x = 90 - 14

x = 76 / 4 = 19

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Find the time required for an investment of $5000 to grow to $8000 at an interest rate of 7% per year if it is compounded quarte
kondaur [170]

Answer:

7 years

Step-by-step explanation:

A = P (1 + r)^(n)

where A is the final amount, P is the initial amount, r is the interest rate, and n is the number of times of compounding.

Compounded quarterly means compounded 4 times per year.  So the effective interest rate per compounding is:

r = 0.07 / 4

r = 0.0175

Given that A = 8000 and P = 5000:

8000 = 5000 (1 + 0.0175)^n

1.6 = 1.0175^n

log 1.6 = n log 1.0175

n = (log 1.6) / (log 1.0175)

n ≈ 27.1

Rounding up to the nearest whole number, it takes 28 compoundings.  Since there's 4 compoundings per year:

t = 28 / 4

t = 7

It takes 7 years.

4 0
4 years ago
This is worth 20 points, please answer! (Please answer with a variable in the equation)
Sindrei [870]

Answer:

The equation is 41 x 3 and 123-10 the answer is 113

Step-by-step explanation:

First, you have to find out how much 3 times is that is 123 then you take away ten which would be 113.

4 0
3 years ago
Please help me with this
Airida [17]

Answer:

1. tangent slope is horizontal at (-5,197) and (1,-19)

2. the tangent line has slope -48 at points (-1, 37) and (-3, 141)

Step-by-step explanation:

1. f(x) = 2x³+12x²-30x-3

f'(x) = 6x²+24x-30

the tangent is horizontal when f'(x) = 0

0 = 6x²+24x-30

0 = x²+4x-5

0 = (x+5)(x-1)

x= -5, 1

f(-5) = -250+300+150-3 = 197

f(1) = -19

tangent slope is horizontal at (-5,197) and (1,-19)

2. when f'(x) = -48

6x²+24x-30 = -48

6x²+24x+18 = 0

x²+4x+3 = 0

(x+1)(x+3) = 0

x = -1, -3

f(-1) = 37

f(-3)=141

the tangent line has slope -48 at points (-1, 37) and (-3, 141)

7 0
3 years ago
What is 0.15 repeating as a fraction
Gre4nikov [31]
.15 repeating as a fraction is 5/33.
8 0
4 years ago
Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
AnnZ [28]

Answer:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

t y'' + 7y = t

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

7 0
4 years ago
Read 2 more answers
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