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3 years ago

Please help solve!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Andrews [41]3 years ago

7 0

Since it passes throug the point
then we can substitute the point because we know it is on the graph
y=ax^2+bx+c substituting (-1,1)
1=a(-1)^2-b+c
1=a-b+c
which is D

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10 points, doing this one again because I got a wrong answer last time but regardless thank you for trying to help

Alik [6]

Hey ! there

Answer:

Value of missing side i.e. TE is 12 feet

Step-by-step explanation:

In this question we are provided with a right angle triangle having TS - 35 ft and SE - 37 ft . And we are asked to find the missing side that is TE using Pythagorean Theorem .

Pythagorean Theorem : -

According to Pythagorean Theorem sum of squares of perpendicular and base is equal to square of hypotenuse in a right angle triangle i.e.

H² = P² + B²

Where ,

H refers to Hypotenuse

P refers to Perpendicular

B refers to Base

Solution : -

In the given triangle ,

Base = TE

Perpendicular = TS ( 35 feet )

Hypotenuse = SE ( 37 feet )

Now applying Pythagorean Theorem :

$\quad \longmapsto \qquad \:SE {}^{2} = TS {}^{2} + TE {}^{2}$

Substituting values :

$\quad \longmapsto \qquad \:37 {}^{2} = 35 {}^{2} + TE {}^{2}$

Simplifying it ,

$\quad \longmapsto \qquad \:1369 = 1225 + TE {}^{2}$

Subtracting 1225 on both sides :

$\quad \longmapsto \qquad \:1369 - 1225 = \cancel{1225} + TE {}^{2} - \cancel{1225}$

We get ,

$\quad \longmapsto \qquad \:144 = TE {}^{2}$

Applying square root to both sides :

$\quad \longmapsto \qquad \ \sqrt{ 144} = \sqrt{TE {}^{2}}$

We get ,

$\quad \longmapsto \qquad \: \red{\underline{\boxed{\frak{TE = 12 \: feet}}}} \quad \bigstar$

Henceforth , value of missing side is 12 feet .

Verifying : -

Now we are verifying our answer using Pythagorean Theorem . We know that according to Pythagorean Theorem ,

SE² = TS² + TE²

Substituting value of SE , TS and TE :

37² = 35² + 12²

1369 = 1225 + 144

1369 = 1369

L.H.S = R.H.S

Hence , Verified .

Therefore , our answer is correct .

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So by definition, area is equal to the length (x) times the width (y). The area of the square mat is = x × y, or xy
If the area of the rectangular mat is twice that of the square mat, the area of the rectangular mat would have to be = 2 × x × y
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