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2 years ago

Question is in the picture

Mathematics

1 answer:

Vinil7 [7]2 years ago

8 0

Answer:

y=1/3x -4

Step-by-step explanation:

3y-x=-12

y= 1/3x-4

m=1/3

(gradient is the same since its parallel)

y=mX+c

(2) = 1/3(18) + c

2 = 6 + c

2 - 6 = c

c= -4

(y-intercept)

Therefore: y=1/3x-4

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(HELP ME 15 PTS)

kap26 [50]

Answer: $\bold{\dfrac{27x^2y^3}{16}}$

<u>Step-by-step explanation:</u>

$\dfrac{(3x^2y^5)^3}{(2xy^3)^4}\\\\\\=\dfrac{3^3\cdot x^{2\cdot 3}\cdot y^{5\cdot 3}}{2^4\cdot x^4\cdot y^{3\cdot 4}}\\\\\\=\dfrac{27\cdot x^6\cdot y^{15}}{16\cdot x^4\cdot y^{12}}\\\\\\=\dfrac{27\cdot x^{6-4}\cdot y^{15-12}}{16}\\\\\\=\dfrac{27\cdot x^2\cdot y^3}{16}$

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2 years ago

A rectangular patio has a length of 12 yards and a width of 8.6 yards.

Savatey [412]

Area = Length x width

Area = 12 x 8.6

Area = 103.2 square yards.

5 0

3 years ago

Read 2 more answers

Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

$\angle KJL = 90^\circ$

Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

6 0

3 years ago

How to find area of this kite?

Y_Kistochka [10]

You can seperate it into two triangles then take the area of each of those (a=1/2bh) and add each triangles area together

7 0

3 years ago

Please help ill give brainlist

nordsb [41]

Answer:

c. m∠2 = m∠5

Step-by-step explanation:

We can check 1 by 1 and find the false one.

For answer A, measures 2 and 4 are vertical angles, which means they are equal, so that's not false.

For answer b, measures 4 and 8 are corresponding angles, since X and Y are parallel, so that's also not false.

For answer c, measures 2 and 5 are supplementary to each other, and since the transversal is not perpendicular to X nor Y, they are not equal, therefore, this is false, making that the answer.

For answer d, measure 2 and 6 are corresponding angles, so that's also not false.

Therefore, the answer must be c.

7 0

3 years ago