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3 years ago

!!PLEASE HELP!!

Mathematics

2 answers:

abruzzese [7]3 years ago

8 0

The water receded 5/6 feet every month

aksik [14]3 years ago

7 0

The water receded 5/6 feet every month

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Hii I need help with these questions tyy!

9966 [12]

Answer:

(7x+2)(3x-1); (5x+4)^2 or (5x+4)(5x+4);

Step-by-step explanation:

formula to find area is L* W = A ; meaning length multiplied by width equals to the area.

21.

using the formula to find area...

l*w=a

(7x+2)(3x-1) would be your answer

btw, don't put the "=a" at the end, since it's an expression :))

22.

since the problem states that it's a square, you can just say either...

(5x+4)^2 or (5x+4)(5x+4) not sure which one your teacher accepts

4 0

2 years ago

Read 2 more answers

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

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

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

$(x-h)^2 + (y-k)^2 = {r}^{2}$

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

3 0

2 years ago

What distance is shorter 3/4 or 2/4

dolphi86 [110]

2/4 is shorter distance. I think. Hope its helps !!!!!!!

8 0

3 years ago

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WILL GIVE BRAINLIEST! Find k so that (5, k) is equidistant from (–1, 2) and (3, 0).

pochemuha

Answer:

Step-by-step explanation:

So first I created 2 distance forumlas and set them equal to each other

$\sqrt{(5+1)^2+(k-2)^2} =\sqrt{(5-3)^2+(k)^2}$

then I simplified them

$(5+1)^2+(k-2)^2 =(5-3)^2+(k)^2\\$

$36 + k^2 - 4k +4 = k^2 + 4$

$-4k + 36 = 0\\$

$4k = 36$

$k = 9$

when k = 9, the two distances are equal

7 0

3 years ago

Please answer this correctly without making mistakes

Mnenie [13.5K]

Answer:

Step-by-step explanation:

750,000,000 times 10^5 can be expressed as 7.5 times $10^{11}$

7.55 times 10^13 is greater than 7.5 times 10^11.

3 0

2 years ago

Read 2 more answers