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2 years ago

Find the perimeter or circumference and area of each figure if each unit on the graph measures

Mathematics

2 answers:

djverab [1.8K]2 years ago

8 0

Answer:

Step-by-step explanation:

It's 20 because you round

lesya692 [45]2 years ago

5 0

Answer:

Below

Step-by-step explanation:

First we can find the length of the two sides that are NOT the hypotenuse :

Counting side AC we can see that AC = 8cm

Counting side BA we can see that BA = 4cm

Now to find the angular side aka the hypotenuse we have to use Pythagorean theorem :

a^2 + b^2 = c^2

Plugging in our values :

(8)^2 + (4)^2 = c^2

64 + 16 = c^2

80 = c^2

Now we have to square this number to get the value of c

√ 80 = c

8.9 = c

Adding all the values up to get the perimeter :

8.9 + 8 + 4 = 20.9 cm

Hope this helps! Best of luck <3

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Can someone help me on this?

n200080 [17]

Answer:

k = -5

Step-by-step explanation:

y = -8x - 73 and x = -4

y = (-8) (-4) - 73 = - 41 : intersect at (-4 , - 41)

(-4 , - 41) must at y = 9x + k

-41 = 9*(-4) +k

k = -41 + 36 = -5

y = 9x - 5 x<-4

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3 years ago

What are the solutions of 2x^2 + 16x + 34 = 0?

nexus9112 [7]

Answer :

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Step-by-step explanation:

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2 years ago

The measure of an angle is 173°. What is the measure of its supplementary angle?

barxatty [35]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has

Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Solving

∆ABC A(11, 6), B(5, 6), and C(5, 17)

AB = 6 units BC = 11 units AC = 12.53 units
∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)
XY = 7.14 units YZ = 18.79 units XZ = 11.66 units

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

MN = 6 units NO = 11 units MO = 12.53 units
∆JKL J(17, -2), K(12, -2), and L(12, 7).
JK = 5 units KL = 9 units JL = 10.30 units
∆PQR P(12, 3), Q(12, -2), and R(3, -2)
PQ = 5 units QR = 9 units PR = 10.30 units
Therefore
we have the ∆ABC and the ∆MNO 
with all three sides equal ---------> are congruent
we have the ∆JKL and the ∆PQR
with all three sides equal ---------> are congruent

let's check

Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

1) If ∆MNO ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC

then ∆MNO ≅ ∆ABC

a) Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

M(-9, -4)-----------------> M1(-9,4)

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

b) Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

∆M1N1O1 M1(-9, 4), N1(-3, 4), and O1(-3, 15).

M1(-9, -4)-----------------> M2(9,4)

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

c) Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

∆M2N2O2 M2(9,4), N2(3,4), and O2(3, 15).

M2(9, 4)-----------------> M3(11,6)=A

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

∆ABC A(11, 6), B(5, 6), and C(5, 17)

∆MNO reflection------- > ∆M1N1O1 reflection---- > ∆M2N2O2 translation -- --> ∆M3N3O3

The ∆M3N3O3=∆ABC

Therefore ∆MNO ≅ ∆ABC - > check list

2) If ∆JKL -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR

then ∆JKL ≅ ∆PQR

d) Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

∆JKL J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)-----------------> J1(2,17)

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

e) translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).

J1(2, 17)-----------------> J2(12,3)=P

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

∆PQR P(12, 3), Q(12, -2), and R(3, -2)

∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR

Therefore ∆JKL ≅ ∆PQR - > check list

6 0

3 years ago

Given a leading coefficient of 8, polynomial roots of 1 & 2, and the known point on the graph (4,5). Write an equation that

m_a_m_a [10]

Given:

The leading coefficient of a polynomial is 8.

Polynomial roots are 1 and 2.

The graph passes through the point (4,5).

To find:

The 3rd root and the equation of the polynomial.

Solution:

The factor form of a polynomial is:

$y=a(x-c_1)(x-c_2)...(x-c_n)$

Where, a is a constant and $c_1,c_2,...,c_n$ are the roots of the polynomial.

Polynomial roots are 1 and 2. So, $(x-1)$ and $(x-2)$ are the factors of the polynomial.

Let the third root of the polynomial by c, then $(x-c)$ is a factor of the polynomial.

The leading coefficient of a polynomial is 8. So, a=8 and the equation of the polynomial is:

$y=8(x-1)(x-2)(x-c)$

The graph passes through the point (4,5). Putting $x=4,y=5$ , we get

$5=8(4-1)(4-2)(4-c)$

$5=8(3)(2)(4-c)$

$5=48(4-c)$

Divide both sides by 48.

$\dfrac{5}{48}=4-c$

$c=4-\dfrac{5}{48}$

$c=\dfrac{192-5}{48}$

$c=\dfrac{187}{48}$

Therefore, the 3rd root on the polynomial is $\dfrac{187}{48}$ .

8 0

2 years ago