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3 years ago

What’s the area of this figure?

Mathematics

1 answer:

Ivan3 years ago

3 0

Answer:

C. 155.5 cm2

Step-by-step explanation:

area = 5*15 + (15+8)*7/2 = 75 + 80.5 = 155.5 cm2

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What is the perimeter of the figure

Brums [2.3K]

Answer:

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. x is in this case the length of the rectangle while y is the width of the rectangle. The perimeter, P, is: P=x+x+y+y.

3 0

3 years ago

Select All That Apply .

makvit [3.9K]

Answer:

the 4th one ,hoped this help

7 0

3 years ago

Please help! College Algebra!

german

The volume would be (x+1)x(x+4)x(3x)= (x^2+x+4x+4)x(3x)= (x^2+5x+4)x(3x)= (3x^3)+(15x^2)+(12x)
Hence, the answer is the 4th option.
Hope this helped you, and if it did, please mark it as the brainiest answer. I would really appreciate it!

4 0

3 years ago

A road race is 13 1/2 miles long. There are water stations every 3/4 if a mile including at the finish line. How

Margarita [4]

The number of water stations there are along the 13 1/2 miles long road race according to the task content is; Choice C; 18 water stations.

<h3>What is the number of water stations along the road race?</h3>

It follows from the task content that the number of water stations can be determined as follows;

13 1/2 ÷ 3/4

= 27/2 × 4/3

= 18 water stations.

Read more on fraction division;

brainly.com/question/407943

#SPJ1

7 0

2 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago