If sin 0 = 6 12 find cos 0 and sin 8.

Write the equation of the line,in slope-intercept form, that has a slope of -2 and goes through the point (1,1)

kirza4 [7]

Answer:

The equation of the line would be y = -2x + 3

Step-by-step explanation:

In order to find the equation of the line, start by using point-slope form with the known information.

y - y1 = m(x -x1)

y - 1 = -2(x - 1)

Now that we have this, solve for y.

y - 1 = -2x + 2

y = -2x + 3

8 0

3 years ago

Joey enter the Hot Dog Eating competition. He ate 30 1 / 2 hot dogs and 3 3 / 4 minutes. How many hot dogs did he eat in one min

VikaD [51]

(30 1/2) / (3 3/4) =
(61/2) / (15/4) =
61/2 * 4/15 =
122/15 =
8 2/15 hot dogs were eaten in 1 minute <==

8 0

3 years ago

Which shapes are quadrilaterals?

lubasha [3.4K]

Hey there! I'm happy to help!

A quadrilateral is any polygon (enclosed shape) with four sides. Let's see what each of these shapes are.

Scalene right triangle - the prefix tri- means three (tricycle, tripod, triple, etc.), and all triangles have three sides, so this is not a quadrilateral.
Obtuse scalene triangle - once again a triangle, so not a quadrilateral.
Isosceles right triangle - would you look at that, another triangle! Not a quadrilateral.
Hexagon - a hexagon is a six-sided polygon (hex=six), so this is not a quadrilateral.
Pentagon - a pentagon is a five-sided polygon (pent=five, like Pentatonix!). Not a quadrilateral
Right trapezoid - a trapezoid is a quadrilateral with at least one pair of parallel sides!
Venn diagram - a Venn diagram is a compare and contrast chart made of two overlapping circles (ZERO sides), so this is not a quadrilateral.

So, the only shape on your list that is a quadrilateral is 6. right trapezoid.

Have a wonderful day! :D

5 0

3 years ago

√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1

Snowcat [4.5K]

$Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}$

$\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s$

7 0

3 years ago

Alex has saved $720, and every month he saves an additional $80. Jax has saved $480, and every month he saves an additional $120

andrew-mc [135]

9514 1404 393

Answer:

more than 10 months

Step-by-step explanation:

Alex's balance starts at 720 and increases by 80 each month. It can be represented by ...

A = 720 +80m

Jax's balance starts at 480 and increases by 120 each month. It can be represented by ...

J = 480 +120m

Jax's balance will be more than 160 more than Alex's when ...

J -A > 160

(480 +120m) -(720 +80m) > 160

40m -240 > 160 . . . . . . . . . . . . . collect terms

m -6 > 4 . . . . . . . . . . . . . . . divide by 40

m > 10 . . . . . . . . . . . . add 6

It will take more than 10 months for Jax to have saved over $160 more than Alex.

8 0

3 years ago