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3 years ago

10. Represent Real-World Problems Describe a real-world situation that

Mathematics

1 answer:

nirvana33 [79]3 years ago

4 0

Answer:

The slope-intercept form of the equation of a line is: y = mx + b

where:

m = slope

b = y-intercept [the place where x=0]

So the line

y = 20x + 0

has

a slope of 20

a y-intercept of 0 [0,0]

it goes through the Origin

Step-by-step explanation:

please mark this answerr the brainkest

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Find the sum of 86331 and 909

devlian [24]

87,240. I added them both together.
Sum= addition

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3 years ago

Which of the following is not a correct description of the graph of the function y = -2x - 7?

swat32

The graph of the function contains the points (0,-7), (1,-9), and (3,-1) is incorrect

it contains all the points except the last set...(3,-1)
y = -2x - 7
-1 = -2(3) - 7
-1 = -6-7
-1 = - 13 (incorrect)

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3 years ago

HELP PLEASE <br> solve the right triangle

aleksandrvk [35]

Answer:

Part 1) $FG=4.4\ units$

Part 2) $EF=3.9\ units$

Part 3) $m\angle G=63^o$

Step-by-step explanation:

step 1

Find the measure of length side FG

In the right triangle EFG

we know that

$sin(27^o)=\frac{GE}{FG}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(27^o)=\frac{2}{FG}$

$FG=\frac{2}{sin(27^o)}=4.4\ units$

step 2

Find the measure of length side EF

In the right triangle EFG

we know that

$cos(27^o)=\frac{EF}{FG}$ ----> by CAH (adjacent side divided by the hypotenuse)

substitute the given values

$cos(27^o)=\frac{EF}{4.4}$

$EF=cos(27^o)(4.4)=3.9\ units$

step 3

Find the measure of angle G

we know that

$m\angle G+27^o=90^o$ ---> by complementary angles in a right triangle

$m\angle G=90^o-27^o=63^o$

7 0

3 years ago

Analyze the scatterplot below, what prediction can you make about the group of students used in the study from the graph below?

SCORPION-xisa [38]

Answer:

i do not understand

Step-by-step explanation:

8 0

3 years ago

A circle is shown. 2 raddi are drawn. A chord connects the points of the radii on the circle to each other to form a triangle. T

iragen [17]

Answer:

$(5\pi-11.6)ft^2$

Step-by-step explanation:

Given:

Length of radius of circle = 5 feets

Length of perpendicular bisector = 4 feets

To find:

Area of the shaded portion of the circle

Solution:

As OD is perpendicular bisector of AB,

$AB=2AD=2(2.9)=5.8\,\,feets$

$\angle ODA=90^{\circ}$

Area of $\Delta AOB$ = 1/2 (base) × (height) = $\frac{1}{2}\times 4\times 5.8=11.6$ square feets

Area of sector AOBC = $\frac{\angle AOB}{360^{\circ}}\pi(r^2)=\frac{72^{\circ}}{360^{\circ}}\pi(5^2)=5\pi$ square feets

Here, r denotes radius of circle

So,

Area of shaded portion = Area of sector AOBC - Area of $\Delta AOB$ = $(5\pi-11.6)ft^2$

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3 0

3 years ago