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3 years ago

the perimeter of a rectangle is 34 units. it’s width is 6.5 units. write the equation to determine the length of the rectangle

Mathematics

1 answer:

zubka84 [21]3 years ago

4 0

Answer:

4 units

Step-by-step explanation:

Perimeter formula is P = 2L + 2W. Here P = 34 units = 2L + 2(6.5 units).

Reducing this, 17 = L + 13. Then L = 4 units.

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Help me. I don't understand. Please help me find X and IK

sertanlavr [38]

Answer:

$x=4\\\\y=6\\\\IK=47$

Step-by-step explanation:

Two Tangent Theorem: Tangents which meet at same point are equal in length.

Here tangents at J and H meet at I.

$Hence\ HI=IJ\\\\y^2-10=4y+2\\\\y^2-4y-12=0\\\\y^2-6y+2y-12=0\\\\y(y-6)+2(y-6)=0\\\\(y-6)(y+2)\\\\y\ can\ not\ be\ negative\ as\ in\ that\ case\ IJ\ will\ be\ negative.\\\\Hence\ y=6\\\\IJ=4y+2=4\times 6+2=26\\\\\\\\Tangents\ at\ J\ and\ L\ meet\ at\ K.\ Use\ two\ tangent\ theorem.\\\\JK=KL\\\\5x+1=6x-3\\\\6x-5x=1+3\\\\x=4\\\\JK=5x+1=5\times 4+1=21\\\\\\IK=IJ+JK=26+21\\\\IK=47$

6 0

3 years ago

Pwease help fast this is mah last question!

inessss [21]

Answer:

A = 222 units^2

Step-by-step explanation:

To find the area of this trapezoid, first draw an imaginary horizontal line parallel to AD and connecting C with AB (Call this point E). Below this line we have the triangle CEB with hypotenuse 13 units and vertical side (21 - 16) units, or 5 units. Then the width of the entire figure shown can be obtainied using the Pythagorean Theorem:

(5 units)^2 + CE^2 = (13 units)^2, or 25 + CE^2 = 169. Solving this for CE, we get |CE| = 12.

The area of this trapezoid is

A = (average vertical length)(width), which here is:

(21 + 16) units

A = --------------------- * (12 units), which simplifies to:

A = (37/2 units)(12 units) = A = 37*6 units = A = 222 units^2

8 0

3 years ago

A line m is perpendicular to an angle bisector of ∠A. The sides of ∠A intersect this line m at points M and N. Prove that △AMN i

Ray Of Light [21]

Answer:

Step-by-step explanation:

Given: A line m is perpendicular to the angle bisector of ∠A. We call this

intersecting point as D. Hence, in figure ∠ADM=∠ADN =90°.

AD is angle bisector of ∠A. Hence, ∠MAD=∠NAD.

To Prove: ΔAMN is an isosceles triangle. i.e any two sides in ΔAMN are

equal.

Solution: Now, In ΔADM and ΔADN

∠MAD=∠NAD ...(1) (∵Given)

AD=AD ...(2) (∵common side)

∠ADM=∠ADN ...(3) (∵Given)

Hence, from equation (1),(2),(3) ΔADM ≅ ΔADN

( ∵ ASA congruence rule)

⇒ AM=AN

Now, In Δ AMN

AM=AN (∵ Proved)

Hence, ΔAMN is an isosceles triangle.

7 0

3 years ago

Help plss i need help

andrew11 [14]

It’s b. you divide by 5 on both sides and at the end you’ll get x= 6

3 0

3 years ago

Find the slope of a line parallel to a line that contains the points (8, -3) and (2, 5).

Illusion [34]

Answer:

$-\frac{4}{3}$ (Your answer choices were really confusing, so I just put the plain answer)

Step-by-step explanation:

To find the slope of a line with just two points, we can use the formula $\frac{y2-y1}{x2-x1}$ . In this case, $\frac{5-(-3)}{2-8}$ . This is $\frac{8}{-6}$ . This reduces to $-\frac{4}{3}$

8 0

2 years ago