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3 years ago

The two longer sides of a triangle Measure 16 and 22 which of the following is a possible length of the shortest side

1 answer:

6 0

The rule for sides of a triangle is that it has to be less than the sum of the other two sides, and more than the difference of the two sides. So, 16 + 22 = 38, which means that it has to be less than 38, and 22 - 16 = 6, which means it has to be more than 6. I don't know the options you're given, but less than 38 and more than 6 are your boundaries. I hope this helps!

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Find four consecutive integers with a sum of 106.

grandymaker [24]

4 consecutive integers...
x, x + 1, x + 2, x + 3

x + x + 1 + x + 2 + x + 3 = 106
4x + 6 = 106
4x = 106 - 6
4x = 100
x = 100/4
x = 25

x + 1 = 25 + 1 = 26
x + 2 = 25 + 2 = 27
x + 3 = 25 + 3 = 28

ur 4 integers are : 25,26,27,28

6 0

3 years ago

What is the surface area of a cube in which each face of the cube has an area of 7 cm^2?

prisoha [69]

Answer:

In a cube, the length and width are the same, so both sides will be 7 cm, yielding an area of 49 square centimeters. The surface area of a 7 cm cube will be 294 cm2 . Hope this helps!

4 0

3 years ago

What's the measure of ABD, DBK, and KBC? Show all work.

Andrew [12]

Answer:

56.6

Step-by-step explanation:

Since BK is a bisector of ABC at B,

this implies that:

ABC = ABK + KBC

ABC = 28.3 + 28.3

ABC= 56.6

5 0

3 years ago

What is the sum of -6 4/5 and 6 4/5

Damm [24]

Answer:

Step-by-step explanation:

same value negative plus positive is 0

8 0

3 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

3 years ago