A triangle has a base of 3 2/5 inches and an area of 5 1/10 square inches. Find the height of the triangle.

Sara bought a used car for $10,000. She learned from online that the value of a car depreciated 15% per year. What is the formul

Fudgin [204]

Answer:

Step-by-step explanation:

Value of Sam's car after t years =

= $=10000*(1-\frac{15}{100})^{t}\\\\=10000*(\frac{85}{100})^{t}\\\\=10000*(0.85)^{t}$

Value of Sam's car in 5 years=

$=10000*(0.85)^{5}\\\\=10000*0.4437\\\\$

= $ 4437

4 0

3 years ago

Yea also all of this

KiRa [710]

Answer:

yea I cant really do this over a screen

5 0

2 years ago

Read 2 more answers

A circle has a radius of 6cm . find the length s of the arc intercepted by a central angle of 169° .

Igoryamba

The circumference of the circle is = 2 x pi x 6 = 37.7cm

To find the arc = 360 divided by 169 = 2.13

The arc = 37.7 divided by 2.13 = 17.69 cm

3 0

3 years ago

The product of 5 and m squared, increased by the sum of the square of m and 5

weqwewe [10]

$The\ product\ of\ 5\ and\ m=5\times m=5m\\\\increased\to+\\\\the\ sum\ of\ a\ square\ of\ m\ and\ 5=m^2+5\\\\\huge\boxed{5m+(m^2+5)=5m+m^2+5}$

8 0

3 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

a)

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

b)

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

3 years ago