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3 years ago

What is the area of this figure?

Mathematics

2 answers:

Anna71 [15]3 years ago

3 0

Hope this can help you

Tamiku [17]3 years ago

3 0

The answer is 142 ft squared.

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A cars velocity on a straight highway increases uniformly from 20km/h to 70km/h in 0.008h. Find the acceleration in km/h2

Bess [88]

Answer:

a = 70 km/h - 20 km/h / 0.008 h = 50 km/h / 0.008 h = 6250 km/h²

3 0

2 years ago

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Help with 5b please . thank you.

Allushta [10]

Answer:

See explanation

Step-by-step explanation:

We are given f(x)=ln(1+x)-x+(1/2)x^2.

We are first ask to differentiate this.

We will need chain rule for first term and power rule for all three terms.

f'(x)=(1+x)'/(1+x)-(1)+(1/2)×2x

f'(x)=(0+1)/(1+x)-(1)+x

f'(x)=1/(1+x)-(1)+x

We are then ask to prove if x is positive then f is positive.

I'm thinking they want us to use the derivative part in our answer.

Let's look at the critical numbers.

f' is undefined at x=-1 and it also makes f undefined.

Let's see if we can find when expression is 0.

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

Find common denominator:

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

(1-1-x+x+x^2)/(1+x)=0

A fraction can only be zero when it's numerator is.

Simplify numerator equal 0:

x^2=0

This happens at x=0.

This means the expression,f, is increasing or decreasing after x=0. Let's found out what's happening there. f'(1)=1/(1+1)-(1)+1=1/2 which means after x=0, f is increasing since f'>0 after x=0.

So we should see increasing values of f when we up the value for x after 0.

Plugging in 0 gives: f(0)=ln(1+0)-0+(1/2)0^2=0.

So any value f, after this x=0, should be higher than 0 since f(0)=0 and f' told us f in increasing after x equals 0.

8 0

2 years ago

Heather brought $7.14 to school. She then spent $1.08 to buy lunch. How much money does she have left?

Wewaii [24]

Answer:

6.06

Step-by-step explanation:

Simple subtraction.

5 0

3 years ago

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Find the general solution of x'1 = 3x1 - x2 + et, x'2 = x1.

Ann [662]

Note that if ${x_2}'=x_1$ , then ${x_2}''={x_1}'$ , and so we can collapse the system of ODEs into a linear ODE:

${x_2}''=3{x_2}'-x_2+e^t$
${x_2}''-3{x_2}'+x_2=e^t$

which is a pretty standard linear ODE with constant coefficients. We have characteristic equation

$r^2-3r+1=\left(r-\dfrac{3+\sqrt5}2\right)\left(r+\dfrac{3+\sqrt5}2\right)=0$

so that the characteristic solution is

${x_2}_C=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}$

Now let's suppose the particular solution is ${x_2}_p=ae^t$ . Then

${x_2}_p={{x_2}_p}'={{x_2}_p}''=ae^t$

and so

$ae^t-3ae^t+ae^t=-ae^t=e^t\implies a=-1$

Thus the general solution for $x_2$ is

$x_2=C_1e^{(3+\sqrt5)/2\,t}+C_2e^{-(3+\sqrt5)/2\,t}-e^t$

and you can find the solution $x_1$ by simply differentiating $x_2$ .

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3 years ago

Find the shaded area.<br> 12 cm<br> 8 cm<br> 8 cm<br> 18 cm

Mrrafil [7]

Answer:

152 cm²

Step-by-step explanation:

The shaded area is calculated as

outer area - inner white area

= (12 × 18) - (8 × 8 )

= 216 - 64

= 152 cm²

7 0

3 years ago

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