What is the perimeter of parallelogram WXYZ

Can someone help me out plz ??!

scoray [572]

Answer:

≈ 62.8

Step-by-step explanation:

Using the equal ratios of

$\frac{areaof circle}{areaofsector}$ = $\frac{circleangle}{sectorangle}$ , that is

$\frac{100\pi }{sector}$ = $\frac{2\pi }{\frac{2\pi }{5} }$ = 5, thus

5 × sector = 100π ( divide both sides by 5 )

sector = 20π = 20 × 3.14 = 62.8 units²

7 0

3 years ago

What is the distance between the points (3, 5) and (3, 1)?

Aleksandr [31]

Answer:

4

Step-by-step explanation:

The x coordinate is the same so the distnce is only the difference in y

5 - 1 = 4

8 0

2 years ago

Read 2 more answers

The area of a rectangle is pool is (15x-9) square units. Factor 15x-9 to find possible dimensions of the pool

Darya [45]

This is the concept of algebra, given that the area of a rectangular pool is (15x-9), the possible dimensions of the pool by factoring will be:
Area=length×width
Area=(15x-9)
factoring the above we get:
Area=3(5x-3)
therefore the possible dimension will be:
length=5x units
width=3 units

8 0

3 years ago

Read 2 more answers

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

2 years ago

Medical researchers have determined that for exercise to be beneficial, a personâ€™s desirable heart rate, r, in beats per minut

Scorpion4ik [409]

Answer:

R, in beats per minute, can be approximated by the formulas, where a represents the person's age.

where R= 143 - 0.65a for women.... (1)

and R= 165 - 0.75a for men,..... (2)

So by implementing these two equation we get a = 40

So the age will be 40 years old

know more about “Mathematical Equation” here: brainly.com/question/1214333

#SPJ4

8 0

1 year ago