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2 years ago

What type of arc is AB

Mathematics

2 answers:

Alinara [238K]2 years ago

6 0

Answer:

B. A minor arc

Step-by-step explanation:

Trust me bro

stiv31 [10]2 years ago

3 0

I'm sorry I don't know the answer but I do have to say that you are loved and no matter how many mistakes you make you will still always be loved. Step-by-step explanation:

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A rectangle has a length of 34 inches less than 5 times it’s width. If the area of the rectangle is 867 square inches, find the

pshichka [43]

Length of the rectangle is 51 inches

Solution:

Given that

Length of the rectangle in 34 miles less than 5 times its width

Area of rectangle = 867 square inches

Need to determine length of the rectangle.

Let assume width of the rectangle be represented by variable x.

So 5 times width of the rectangle =5x

34 inches less than 5 times width of the rectangle = 5x – 34

As Length of the rectangle in 34 miles less than 5 times its width

=> Length of the rectangle = 5x – 34

The formula for rectangle is given as:

$\text { area of rectangle }=length \times width$

As given that Area of rectangle = 867 square inches

$\begin{array}{l}{=>867=(5 x-34) \times(x)} \\\\ {=>5 x^{2}-34 x-867=0}\end{array}$

On solving quadratic equation by using quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case a = 5, b = -34 and c = -867

On substituting value of a, b and c in quadratic formula we get

$\begin{array}{l}{x=\frac{-(-34) \pm \sqrt{(-34)^{2}-4 \times 5 \times(-867)}}{2 \times 5}} \\\\ {=>x=\frac{34 \pm \sqrt{18496}}{10}=\frac{34+136}{10}} \\\\ {=>x=\frac{34+136}{10}=17 \text { or } x=\frac{34-136}{10}=-10.2}\end{array}$

Since x represents width, it cannot be negative, so x = 17

Length of the rectangle = 5x – 34 = 5(17) – 34 = 51

Hence length of the rectangle is 51 inches

5 0

3 years ago

Whats the vertex at of y-13=2(x+4)^2

Kipish [7]

The vertex form:
$y=a(x-h)^2+k$
(h,k) - the coordinates of the vertex

$y-13=2(x+4)^2 \\
y=2(x+4)^2+13 \\
y=2(x-(-4))^2+13 \\ \\
h=-4 \\ k=13$

The coordinates of the vertex are (-4,13).

7 0

3 years ago

Read 2 more answers

Cubic yards of concrete will be required for 14 parking lot light pole standards

riadik2000 [5.3K]

Using conversion of units and volume of a cylinder, it is found that 23.17 cubic yards of concrete are needed, option D.

The volume of a cylinder of radius r and height h is given by:

$V = \pi r^2h$

In this problem:

We want the volume in cubic yards, thus, the units of radius and height have to be in yards.
32 inches in diameter, thus 16 inches of radius. Since each inch has 0.0278 yards, this is a radius of 0.44 yards, thus $r = 0.44$ .
Height of 8 feet, and since each feet has 0.33 yards, this is a height of 2.67 yards, thus $h = 2.67$
14 poles, thus the volume is multiplied by 14.