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2 years ago

7

Graph the solution to the inequality on the number line m >-3

1 answer:

Mashutka [201]2 years ago

7 0

Answer:

sorry I can't remember this

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In the opposite figuret triangle OAB has an area of 72 cm2 and. Triangle ODC has an area of 288 Cm2 Then XandY equal

Lisa [10]

area of OAB = 72 = (1/2) sin (AOB) * OA * OB solve the above for sin(AOB) to find sin(AOB) = 1/2 area of ODC = 288 = (1/2) sin (DOC) * OD * OD Note that sin(DOC) = sin(AOB) = 1/2, OD = 18 + y and OC = 16 + x and substitute in the above to obtain the first equation in x and y 1152 = (18 + y)(16 + x)...

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3 0

3 years ago

in a classroom 5/8 of the students are wearing blue shirts and 1/4 for wearing white shirts there are 24 students in the classro

Oliga [24]

Answer:

3

Step-by-step explanation:

Those wearing a shirt of another color are ...

1 - 5/8 -1/4 = 8/8 -5/8 -2/8 = 1/8

of the total number of students in the classroom

(1/8)×(24 students) = 3 students . . . . wearing another color

_____

<em>Alternate solution</em>

With the given information, you know ...

(5/8)(24) = 15 . . . students wear blue

(1/4)(24) = 6 . . . . students wear white

24 -15 -6 = 3 . . . students wear another color

6 0

3 years ago

Please help ! assignment is due at 11:59 pm !

I am Lyosha [343]

Answer:

sorry hindi ako nakahabol

6 0

3 years ago

Suppose lines EF and GH are reflected over the line Y equals X to form the lines JK and LM which statement would be true about t

katrin [286]

Answer:

A) JK and LM will be parallel to each other.

Step-by-step explanation:

On reflection on $y=x$ line the x co-ordinate changes with y co-ordinate and y co-ordinate changes with x co-ordinate

$(x,y)\rightarrow (y,x)$

Points on line EF

$(0,6) , (-5,-2)$

On reflection of this line on $y=x$ the new points we get for line JK are

$(6,0),(-2,-5)$

Points on line GH

$(-4,9),(-9,1)$

On reflection on y=x line the new points we get for line LM are

$(9,-4),(1,-9)$

Slope of line JK

$m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-5)-0}{(-2)-6} \\m=\frac{-5}{-8}=\frac{5}{8}$

Slope of line LM

$m=\frac{y_2-y_1}{x2-x1}\\m=\frac{(-9)-(-4)}{1-9} \\m=\frac{-9+4}{-8}=\frac{-5}{-8}\\m=\frac{5}{8}$

For two line to be parallel, their slopes will be same.

$m_{JK} =\frac{5}{8} , m_{LM}=\frac{5}{8}$

Since slopes of lines JK and LM are same therefore we can say that these are parallel to each other.

5 0

3 years ago

(5x10,000) +(2x100) +(9x1) +(3x1/1000)

zubka84 [21]

Scientific notation:
5.0209003 x 10^4

7 0

3 years ago