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3 years ago

Solve........................

Mathematics

1 answer:

docker41 [41]3 years ago

3 0

Hello,

here is the picture:

Slope (PR):

$m=\dfrac{2-4}{6-2} =-\dfrac{1}{2} \\Slope\ of\ the\ perpendicular: -\dfrac{1}{\dfrac{-1}{2} } =2\\\\perpendicular\ is\ passing\ trought\ M(4,3): y-3=2(x-4)\\y=2x-5\\If\ y=0\ then\ x= \dfrac{5}{2} \\Q=( \dfrac{5}{2},0)\\$

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Graduate management aptitude test (gmat) scores are widely used by graduate schools of business as an entrance requirement. supp

klemol [59]

The values that exist two standard deviations above and below the mean are 681 and 265.

<h3>How to determine the values?</h3>

The given parameters exist:

Mean = 473

Standard deviation =104

The values above and below the mean exist calculated utilizing:

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In this case n = 2.

So, we have, $x=\bar{x} \pm 2 \sigma$

The value above is:

x = 473 + 2 $*$ 104 = 681

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2 years ago

Solve the following system using the SUBSTITUTION method.<br> x - 3y = 2<br> 2x - 6y = 6

notka56 [123]

ANSWER:

x – 3y = 2 ...(1)
2x – 6y = 6 ...(2)

On solving eq(1), we get

x – 2 = 3y
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Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

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

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago

Solve........................​

Solve........................