All categories

3 years ago

PLEASE help me with this question with solutions

Mathematics

1 answer:

ss7ja [257]3 years ago

8 0

Answer:

Step-by-step explanation:

the inscribed angle intersects an arc that twice the measure of the arc interested by the central angle. The inscribed angle's arc measures 144° and the central angle's arc measures 72°

>central angle and the arc the intersected arc are equal i.e they are both 72°

>the inscribed angle is half of the measure of the intersected arc i.e since the inscribed angle is 72° the arc interested is 144°

You might be interested in

Hiroto reduces a rectangular photograph by 0.5 by scanning it onto his computer. The image on the screen measures 3 inches by 5

AnnyKZ [126]

Answer:

The correct option will be: Rectangle C (Width= 6 and Length= 10)

Step-by-step explanation:

Suppose, the length of the original photograph is $x$ inches and width is $y$ inches.

Hiroto reduces the photograph by 0.5 by scanning it onto his computer.

So, the length of the scanned image $=0.5x$ inches and width $=0.5y$ inches.

Given that, the length and width of the image on the screen are 5 inches and 3 inches respectively. That means.........

$0.5x=5\\ \\ x=\frac{5}{0.5}=10\\\\\\ and\\ \\ 0.5y=3\\ \\ y=\frac{3}{0.5}=6$

So, the rectangle $C$ represents the original photograph, which has width as 6 inches and length as 10 inches.

4 0

3 years ago

Read 2 more answers

The slope of the line below is - 8/7 Write a point-slope equation of the line

julsineya [31]

Answer:

O C. y - 4 = - 8/7(x - 4)

Step-by-step explanation:

https://www.albert.io/blog/point-slope-form/#What_is_point_slope_form

7 0

3 years ago

: Find the approximate perimeter of the isosceles triangle A OPD.

AleksAgata [21]

Answer:

24.98 units

Step-by-step explanation:

The picture of the question in the attached figure

we have the coordinates

P(1,-6),D(4,3),O(7,-6)

The perimeter of triangle OPD is equal to

$P=OP+PD+OD$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance OP

we have

O(7,-6),P(1,-6)

substitute in the formula

$d=\sqrt{(-6+6)^{2}+(1-7)^{2}}$

$d=\sqrt{(0)^{2}+(-6)^{2}}$

$d_O_P=6\ units$

step 2

Find the distance PD

we have

P(1,-6),D(4,3)

substitute in the formula

$d=\sqrt{(3+6)^{2}+(4-1)^{2}}$

$d=\sqrt{(9)^{2}+(3)^{2}}$

$d_P_D=\sqrt{90}\ units$

step 3

Find the distance OD

we have

O(7,-6),D(4,3)

substitute in the formula

$d=\sqrt{(3+6)^{2}+(4-7)^{2}}$

$d=\sqrt{(9)^{2}+(-3)^{2}}$

$d_O_D=\sqrt{90}\ units$

step 4

Find the perimeter

$P=6+\sqrt{90}+\sqrt{90}$

$P=6+9.49+9.49=24.98\ units$

4 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

W hich of the following are you not likely to see on a graph? dependent variable, units, independent variable, prediction ...?

nekit [7.7K]

"Prediction" is the one among the following choices given in the question that you are not likely to see on a graph. The correct option among all the options that are given in the question is the fourth option or the last option. I hope that this is the answer that has actually come to your desired help.

3 0

3 years ago

Read 2 more answers