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3 years ago

Please help.. I got stuck on this question please help and explain

Mathematics

2 answers:

Inessa [10]3 years ago

8 0

Jay is wrong about his claim

Deffense [45]3 years ago

8 0

150 + 130 = 280 combined
they donated 1/4 of the total amount....1/4(280) = 280/4 = 70....so now they have 280 - 70 = 210
then they spent 40%.....210 - 0.4(210) = 210 - 84 = 126..what they have left

35% of 280 = 0.35(280) = 98

Jay is not correct

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Mario invested $5100 at 13%

cestrela7 [59]

Answer:

Amount = $6614 and 19 cent

Step-by-step explanation:

Formula for compound interest is

A= p(1+r/n)^(nt)

Compounded daily

So n= 365*2= 730

T= 2

r= 0.13

P= 5100

A= p(1+r/n)^(nt)

A= 5100(1+0.13/730)^(730*2)

A= 5100(1+1.78082*10^-4)^(1460)

A= 5100(1.000178082)^1460

A= 5100(1.2969)

A= 6614.19

Amount = $6614 and 19 cent

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=4%20%20%5Ccos%2860%20%2B%20%20%5Calpha%20%29%20%20%5Csin%2830%20%2B%20%20%5Calpha%20%29%20%20%

Yuri [45]

Answer:

i am very bad at o math i am only at 8 sorry

Step-by-step explanation:

if you could give me a brainliest.

5 0

3 years ago

-3+ 3x (x-7)=-6 (-5+ 4×)

Dennis_Churaev [7]

The answer to this question using the quadratic formula is x= -1+3√5/2 and x=-1-3√5/2
that is a square root, or, 3 times the square root of 5 ALL over 2, except for the 1. Same with the other one.

8 0

3 years ago

The slope-intercept form of the equation of a line that passes through point (-2, -13) is y = 54 -3. What is the point slope

dangina [55]

Answer:

$y+13=5(x+2)$

Step-by-step explanation:

The correct question is

The slope-intercept form of the equation of a line that passes through point (-2, -13) is y = 5x -3. What is the point slope form of the equation for this line?

we have

$y=5x-3$

This is the equation of the line in slope intercept form

where

the slope is $m=5$

the y-intercept is $b=-3$

Remember that

The equation of the line in point-slope form is equal to

$y-y1=m(x-x1)$

we have

$m=5$

$point\ (-2,-13)$

substitute

$y-(-13)=5(x-(-2))$

$y+13=5(x+2)$

3 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago