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3 years ago

5

U th th V 3 D F Geometry

1 answer:

loris [4]3 years ago

8 0

Answer:

interesting can i get brainliest

Step-by-step explanation:

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Y=3/8x−4 write in standard form

Naddik [55]

Answer:

y=3/8x-4
Move the expression to the left
y-3/8x=-4
Multiply both sides by 8
8y-3x=-32
Reorder the terms
-3x+8y=-32
Change the signs and standard form is...
3x-8y=32

5 0

3 years ago

Read 2 more answers

a video game arcade offers a yearly membership with reduced rates for game play. A single membership costs $60 per year. Game to

7nadin3 [17]

slope is $0.10 <em>($1.00 per 10 tokens = $0.10)</em>

y-intercept is $60 <em>($60 is the annual fee)</em>

y = .10x + 60 <em>(y = mx + b)</em>

domain is x ≥ 0 <em>(you can't buy a negative number of token)</em>

range is y ≥ 60 <em>(input the domain (x ≥ 0) to find the range)</em>

Answer: the y-intercept of the function is $60

6 0

3 years ago

Read 2 more answers

A figure is composed of a rectangle and a right triangle. What is the area of the figure?

strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

$A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h$

We substitute the values to obtain;

$Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm$

$Area=\frac{1}{2}(21.6cm)\times 6cm$

$Area=64.8cm^2$

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

$C=2\pi r$

where $r=3.5ft$ is the radius of the circular table cloth.

We substitute this value and $\pi=3.14$ into the formula to get;

$C=2(3.14)(3.5)ft$

$C=21.98ft$

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

$C=\pi r$

where $r=12.5cm$ is the radius of the semicircle.

$C=3.14\times 12.5cm$

$C=39.25cm$

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

$Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh$

$Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24$

$Area=345.3125+300$

$Area=645.3125cm^2$

$Area\approx645cm^2$

QUESTION 6

The area of the parallelogram is given by the formula;

$A=bh$

From the diagram, $b=y\:in.,h=3in.$ .

Given, A=23.7 inches squared.

We substitute the values to obtain;

$23.7=3y$

$y=\frac{23.7}{3}$

$y=7.9in.$

QUESTION 7

The area of a rectangle is given by the formula;

$Area=l\times b$

The area of the bigger rectangle $=9\times15=135in^2$

The area of the smaller rectangle $=8\times 13=104in^2$

The area of the shaded region $=135-104=31in^2$

5 0

3 years ago

In 1990, California switched from a 6/49 lottery to a 6/53 lottery. Later, the state switched again to a 6/51 lottery. (a) Find

zalisa [80]

Answer:

(a) 9.9321*10^{-11}

(b) 6.0498*10^{-11}

(c) 7.7120^{-11}

(d) 1.6417 times more probable

Step-by-step explanation:

(a) The probability of winning a 6/49 lottery is given by:

$P = \frac{1}{49*48*47*46*45*44} =9.9321*10^{-11}$

(b) The probability of winning a 6/53 lottery is given by:

$P = \frac{1}{53*52*51*50*49*48} =6.0498*10^{-11}$

(c) The probability of winning a 6/51 lottery is given by:

$P = \frac{1}{51*50*49*48*47*46} =7.7120^{-11}$

(d) The ratio between the probabilities of winning the 6/49 lottery and winning the 6/53 lottery is:

$r=\frac{9.9321*10^{-11}}{6.0498^{-11}}\\r=1.6417$

It is 1.6417 times more probable.

8 0

3 years ago

Paraphin [41]

Answer:

Where’s the graph?

Step-by-step explanation:

3 0

2 years ago