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2 years ago

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1: In Rectangle ABCD, AC is 5x - 12, and BD is 2x + 15. Find x.

1 answer:

Mice21 [21]2 years ago

8 0

Answer:

X=9

Step-by-step explanation:

In the picture above.

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Solve each equation with the quadratic formula. 4n^2+7n-1=-4

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The answer i got was
n=-3/4,n=-1

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2 years ago

Find m∠Y if m∠Y is six more than three times its supplement.

Svetlanka [38]

To answer this item, we let x be the measure of angle Y. If we say supplement, it is understood that when we sum up the measures of the two angles (angle Y and its supplement) the answer should be 180. Therefore the other angle is 180-x. Going back to the condition given in this item, we will be able to generate an equation,
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2 years ago

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7) Newton sees 50 frogs in the morning, and 7 fewer frogs in the

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Answer:

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Step-by-step explanation:

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2 years ago

Find the distance between the points with the given coordinates (-8,-6),(4,10)

Serggg [28]

Hello from MrBillDoesMath!

Answer:

20

Discussion:

Let (x1,y1) = (-8, -6) and (x2,y2) = (4,10). Per the distance formula,

Distance = sqrt ( (x1-x2)^2 + (y1-y2)^2 ) so

Distance = sqrt ( (-8-4)^2 + (-6-10)^2 )

= sqrt( (12)^2 + (-16)^2 )

= sqrt( 144 + 256 )

= sqrt(400)

= 20

Thank you,

MrB

3 0

2 years ago

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HELPPPP<br> please help me answer this question-

erik [133]

Answer:

$\displaystyle GH = 2 \sqrt{30}$

$\displaystyle FH = 2 \sqrt{10}$

$\displaystyle m \angle G = {30}^{ \circ}$

Step-by-step explanation:

<h3>QUESTION-2:</h3>

we are given a right angle triangle

it's a 30-60-90 triangle of which FH is the shortest side

remember that,in case of 30-60-90 triangle the the longest side is twice as much as the shortest side thus

our equation is

$\displaystyle 2FH=4\sqrt{10}$

divide both sides by 2

$\displaystyle\frac{2FH}{2}= \frac{ 4 \sqrt{10} }{2}$

$\displaystyle FH =2 \sqrt{10}$

<h3>Question-1:</h3>

in order to figure out GH we can use Trigonometry because the given triangle is a right angle triangle

as we want to figure out GH we'll use sin function

remember that,

$\displaystyle \sin( \theta) = \frac{opp}{hypo}$

let our opp, hypo and $\theta$ be GH, 4√10 and 60° respectively

thus substitute:

$\displaystyle \sin( {60}^{ \circ} ) = \frac{GH}{4 \sqrt{10} }$

recall unit circle:

$\displaystyle \frac{ \sqrt{3} }{2} = \frac{GH}{4 \sqrt{10} }$

cross multiplication:

$\displaystyle 2 GH = 4 \sqrt{10} \times \sqrt{3}$

simplify multiplication:

$\displaystyle 2 GH = 4 \sqrt{30}$

divide both sides by 2:

$\displaystyle GH = 2 \sqrt{30}$

<h3>QUESTION-3:</h3>

Recall that, the sum of the interior angles of a triangle is 180°

therefore,

$\displaystyle m \angle G + {60}^{ \circ} + {90}^{ \circ} = {180}^{ \circ}$

simplify addition:

$\displaystyle m \angle G + {150}^{ \circ} = {180}^{ \circ}$

cancel 150° from both sides

$\displaystyle m \angle G = {30}^{ \circ}$

6 0

2 years ago

Read 2 more answers