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3 years ago

How do I take picture on this app

Mathematics

2 answers:

Katena32 [7]3 years ago

8 0

You're home page I think

NemiM [27]3 years ago

3 0

When you r typing ur question, there is an Ask Question button, click that then click on the camera icon to take pic or gallery icon to pick pic from gallery

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I NEED HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

sashaice [31]

Answer:

a,b,b

Step-by-step explanation:

because corellation does not mean causation

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3 years ago

[10÷(7+3)]3 PLEASE HELP

makkiz [27]

Answer:

Step-by-step explanation:

add perenthasis (10 \ 10) 3

(10\10) is 1

(1)3 = 3

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2 years ago

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Alinara [238K]

More info please :):):):):):):)

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3 years ago

Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-7, -5), B(-

sergey [27]

The length of side CD is equal to 11 units.

Given coordinates of rectangle are A(-7,-5), B(-7,6),C(-4,6) ,D(-4,-5)

We are require to find the length of side CD.

Rectangle is a two dimensional figure whose two opposite sides are equal to each other. Distance between two coordinates is the equal to length of a side.

We know that the distance means how far two points are located.

Coordinates of A(-7,-5), B(-7,6),C(-4,6) ,D(-4,-5).

Distance= $\sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2} }$ where ( $x_{1} ,y_{2} )(x_{2} ,y_{2} )$ are the coordinates of two ends of a line.

CD= $\sqrt{(-5-6)^{2} +(-4+4)^{2} }$

= $\sqrt{11^{2}+0^{2} }$

= $\sqrt{121}$

=11 units.

Hence the length of side CD is 11 units.

Learn more about distance at brainly.com/question/17273444

#SPJ4

Question is wrong as it should be :

Coordinates of rectangle A(-7,-5), B(-7,6),C(-4,6) ,D(-4,-5) and find the length of CD.

8 0

2 years ago

Can someone confirm this? is it correct?

EastWind [94]

<span>60 Sorry, but the value of 150 you entered is incorrect. So let's find the correct value. The first thing to do is determine how large the Jefferson High School parking lot was originally. You could do that by adding up the area of 3 regions. They would be a 75x300 ft rectangle, a 75x165 ft rectangle, and a 75x75 ft square. But I'm lazy and another way to calculate that area is take the area of the (300+75)x(165+75) ft square (the sum of the old parking lot plus the area covered by the school) and subtract 300x165 (the area of the school). So (300+75)x(165+75) - 300x165 = 375x240 - 300x165 = 90000 - 49500 = 40500 So the old parking lot covers 40500 square feet. Since we want to double the area, the area that we'll get from the expansion will also be 40500 square feet. So let's setup an equation for that: (375+x)(240+x)-90000 = 40500 The values of 375, 240, and 90000 were gotten from the length and width of the old area covered and one of the intermediate results we calculated when we figured out the area of the old parking lot. Let's expand the equation: (375+x)(240+x)-90000 = 40500 x^2 + 375x + 240x + 90000 - 90000 = 40500 x^2 + 615x = 40500 x^2 + 615x - 40500 = 0 Now we have a normal quadratic equation. Let's use the quadratic formula to find its roots. They are: -675 and 60. Obviously they didn't shrink the area by 675 feet in both dimensions, so we can toss that root out. And the value of 60 makes sense. So the old parking lot was expanded by 60 feet in both dimensions.</span>

8 0

3 years ago